Introduction - National Institute of Technology Karnataka · 2.R. Courant and F.John, Introduction to Calculus and Analysis, Volume II, Springer-Verlag 3.N. Piskunov, Di erential

Introduction

P. Sam Johnson

August 7, 2019

P. Sam Johnson Introduction August 7, 2019 1/288

About the Course and Instructor

Course Code : MA110

Course Title : Engineering Mathematics - 1

L-T-P : 3-0-0

Credits : 3

Course Instructor : Dr. P. Sam Johnson

Email : [email protected]

Website : sam.nitk.ac.in

Department : Mathematical and Computational Sciences


Evaluation Plan

25 % Weightage of Quizzes and Classroom Activities

25 % Weightage of Mid-sem

50 % Weightage of End-sem


Objective of the course

To expose the students to the calculus of two or more variables

limits, continuity, partial differentiation and differentiability

concepts of surface, area, volume through multiple integrals

line integrals and eventually Stokes’ and divergence theorems.

Skill development of the student expected from the course

Understanding the concept of limits, differentiability and integration forfunctions of two or more variables ; capability to analyze and solveproblems in integral calculus and vector calculus effectively.


Contents

Functions of Several Variables : Definition, Region in a Plane, Level Curves, LevelSurfaces, Limits, Continuity, Partial Derivatives, Differentiability, Gradients, DirectionalDerivatives, Normals to Level Curves and Tangents, Extreme Values and Saddle Points,Lagrange Multipliers. (Sections 14.1-14.3, 14.5-14.8 ; [1]).

Integral Calculus : Double Integrals, Areas, Double Integrals in Polar Form, TripleIntegrals in Rectangular Coordinates, Triple integrals in Cylindrical and Sphericalcoordinates, Substitutions in Multiple Integrals. (Sections 15.1-15.4, 15.6-15.7; [1]).

Vector Calculus : Line Integrals (exclude mass and moment calculations), Vector Fields,Work, Circulation and Flux, Path Independence, Potential Functions, and ConservativeFields, Greens Theorem in the Plane, Surface Area and Surface Integrals, (exclude massand moments of thin shells), Parametrized Surfaces, Stokes theorem (without proof), TheDivergence theorem (without proof), (exclude Gauss’s Law of Electromagnetic Theoryand the Continuity Equation of Hydrodynamics). (Sections 16.1-16.8; [1]).


Reference Books

1. M.D. Weir, J. Hass and F.R. Giordano, Thomas’ Calculus, 11thEdition, Pearson Publishers.

2. R. Courant and F.John, Introduction to Calculus and Analysis,Volume II, Springer-Verlag

3. N. Piskunov, Differential and Integral Calculus, Vol I & II (Translatedby George Yankovsky).

4. E. Kreyszig, Advanced Engineering Mathematics, Wiley Publishers.


Limits and Continuity

The concept of a limit is fundamental to finding the velocity of a movingobject and the tangent to a curve. We first discuss limits of functions ofsingle variable. We use limits to describe the way a function varies.

Some functions vary continuously; small changes in x produce onlysmall changes in f (x).

Other functions can have values that jump, vary erratically, or tend toincrease or decrease without bound.

The notion of limit gives a precise way to distinguish between thesebehaviors. Also limits play a major role in calculus and the study ofchange.


Average Rates of Change and Secant Lines

We discuss the average rate of change of a function y = f (x) with respectto x over the interval [x1, x2].

Given an arbitrary function y = f (x), we calculate the average rate ofchange of y with respect to x over the interval [x1, x2] by dividing thechange in the value of y , ∆y = f (x2)− f (x1), by the length∆x = x2 − x1 = h of the interval over which the change occurs.

Definition 1.

The average rate of change of y = f (x) with respect to x over theinterval [x1, x2] is

∆y

∆x=

f (x2)− f (x1)

x2 − x1=

f (x1 + h)− f (x1)

h, h 6= 0.



Geometrically, the rate of change of f over [x1, x2] is the slope of the linethrough the point P(x1, f (x1)) and Q(x2, f (x2)).

The slope (gradient) of a line is a number that describes both thedirection and the steepness of the line. Slope is calculated by finding theratio of the “vertical change” to the “horizontal change” between (any)two distinct points on a line.

A secant to the graph y = f (x).

Its slope is ∆y/∆x , the average rateof change of f over the interval[x1, x2].



Let’s consider what happens as the point Q approaches the point P alongthe curve, so the length h of the interval over which the change occursapproaches zero.

A secant to the graph y = f (x).

Its slope is ∆y/∆x , the average rateof change of f over the interval[x1, x2].

In geometry, a line joining two points of a curve is a secant to the curve.Thus, the average rate of change of f from x1 to x2 is identical withthe slope of secant PQ.


Application

Experimental biologists often want to know the rates at which populationsgrow under controlled laboratory conditions.

Example 2 (The Average Growth Rate of a LaboratoryPopulation).

The following figure shows how a population of fruit flies grew in a 50-dayexperiment. The number of flies was counted at regular intervals, thecounted values plotted with respect to time, and the points joined by asmooth curve. Find the average growth rate from day 23 to day 45.


Solution

There were 150 flies on day 23 and 340 flies on day 45. Thus the numberof flies increased by 340− 150 = 190 in 45− 23 = 22 days.

The average rate of change of the population from day 23 to day 45 is∆p∆t = 190

22 ≈ 8.6 flies/day.

This average is the slope of the secant through the points P and Qon the graph.


Growth Rate

The average rate of change from day 23 to day 45 calculated in Example 2does not tell us how fast the population was changing on day 23itself.

For that we need to examine time intervals closer to the day in question.

Example 3 (The Growth Rate on Day 23).

How fast was the number of flies in the population of Example 2 growingon day 23 ?


Solution

To answer this question, we examine the average rates of change overincreasingly short time intervals starting at day 23.

In geometric terms, we find these rates by calculating the slopes of secantsfrom P to Q, for a sequence of points Q approaching P along the curve.


Solution (contd...)

We can observe the following from the table and the graph.

1. As the t-coordinate of Q decreases from 45 to 30, the secant slopesrise from 8.6 to 16.4. Hence we would expect the slopes to riseslightly higher as t continued on toward 23.

2. Geometrically, the secants rotate about P and seem to approach thered line in the figure, a line that goes through P in the same directionthat the curve goes through P. We will see that this line is called thetangent to the curve at P.


Solution (contd...)

Since the line appears to pass through the points (14, 0) and (35, 350), ithas slope

350− 0

35− 14= 16.6 flies/day (approximately).

Thus, on day 23, the population was increasing at a rate of about16.7 flies/day.


Exercises

Exercises 4.

Find the average rate of change of the function over the given interval orintervals.

1. f (x) = x3 + 1

(a) [2, 3] (b) [−1, 1]

2. h(t) = cot t

(a) [π/4, 3π/4] (b) [π/6, π/2]

3. R(θ) =√

4θ + 1 ; [0, 2]

4. P(θ) = θ3 − 4θ2 + 5θ ; [1, 2]


Solutions

1. (a) 28−93−2 = 19 (b) 2−0

1+1 = 1

2. (a) −1−1π/2 = −4

π (b) 0−√

3π/3 = −3

√3

π

3. 3−12 = 1

4. 2− 2 = 0.


Slope of the line

Geometrically, the rate of change of f over [x1, x2] is the slope of the linethrough the point P(x1, f (x1)) and Q(x2, f (x2)).

The slope (gradient) of a line is a number that describes both thedirection and the steepness of the line. Slope is calculated by finding theratio of the “vertical change” to the “horizontal change” between (any)two distinct points on a line.

The slope of a straight line tells us the rate at which it rises or falls.

If there is a tangent line to the curve at P (a line that just touches thecurve like the tangent to a circle), it would be reasonable to identify theslope of the tangent as the slope of the curve at P.


Tangency for circles

A line L is tangent to a circle at a point P if L passes through Pperpendicular to the radius at P.

Such a line just touches the circle.


Tangency for general curves

The following steps tell us an approach that takes into account thebehavior of the secants through P and nearby points Q as Q movestoward P along the curve.

1. Start with what we can calculate, namely the slope of the secant PQ.

2. Investigate the limiting value of the secant slope as Q approaches Palong the curve.

3. If the limit exists, take it to be the slope of the curve at P and definethe tangent to the curve at P to be the line through P with this slope.


Tangency for general curves

The following figure illustrates the geometric idea for the tangent to acurve.

The tangent to the curve at P is the line through P whose slope is thelimit of the secant slopes as Q moves towards P along the curve fromeither side (denoted by Q → P).


Example

Example 5.

Find the slope of the parabola

y = x2

at the point P(2, 4). Write an equation for the tangent to the parabola atthis point?


Solution

We begin with a secant line through P(2, 4) and Q(2 + h, (2 + h)2) nearby.

Secant slope =∆y

∆x=

(2 + h)2 − 22

h= h + 4.

If h > 0, then Q lies above and to the right of P.

If h < 0, then Q lies to the left of P.

In either case, as Q approaches P along the curve, h approaches zero andthe secant slope h + 4 approaches 4. We take 4 to be the parabola’s slopeat P.


Solution (contd...)

Tangent Line :

The tangent to the parabola at P is the line through P with slope 4:

y − 4 = 4(x − 2)

hence y = 4x − 4.


Exercises

Exercise 6.

Use the method in Example 5 to find

(a) the slope of the curve at the given point P, and

(b) an equation of the tangent line at P.

1. y = x2 − 3, P(2, 1)

2. y = x2 − 4x , P(1,−3)

3. y = x3 − 3x2 + 4, P(2, 0).


Solution

1. (a) Slope is 4(b) y = 4x − 7

2. (a) Slope is −2(b) y = −2x − 1

3. (a) Slope is 0(b) y = 0


Limiting of Function Values

Let f (x) be defined on an open interval about x0, except possibly at x0

itself.

If f (x) gets arbitrarily close to L (as close to L as we like) for all xsufficiently close to x0, we say that f approaches the limit L as xapproaches x0, we write

limx→x0

f (x) = L,

which is read “the limit of f (x) as x approaches x0 is L′′.

Essentially, the definition says that the values of f (x) are close to thenumber L whenever x is close to x0 (on either side of x0).


“Informal” Definition of Limit

This definition is “informal” because phrases like arbitrarily close andsufficiently close are imprecise ; their meaning depends on the context.

To a machinist manufacturing a piston, close may mean within a fewthousandths of an inch.

To an astronomer studying distant galaxies, close may mean within a fewthousand light-years.

To a mathematician, “close” may mean something in ε (epsilon) and δ(delta) which, in general, students don’t like to use them.


Behavior of a Function Near a Point

Example 7.

How does the function

f (x) =x2 − 1

x − 1

behave near x = 1 ?


Solution

The given formula defines f for all real number x except x = 1 (we cannotdivide by zero).

For any x 6= 1, we can simplify the formula by factoring the numerator andcanceling common factors :

f (x) = x + 1 for x 6= 1.


Solution (contd...)

The graph of f is thus the line y = x + 1 with the point (1, 2) removed.

The removal point is shown as a “hole” in the figure. Even though f (1) isnot defined, it is clear that we can make the value of f (x) as close as wewant to 2 by choosing x close enough to f .

The graph of f is identical with the line y = x + 1 except at x = 1, wheref is not defined.


Solution (contd...)

We say that f (x) approaches the limit 2 as x approaches 1, and write

limx→1

f (x) = 2, or limx→1

x2 − 1

x − 1= 2.


The limit value does not depend on how the function isdefined at x0

f has limit 2 as x → 1 eventhough f is not defined atx = 1.

g has limit 2 as x → 1 eventhough 2 6= g(1).

h has limit 2 as x → 1 whichequals the value of h atx = 1.

The limits of f (x), g(x), and h(x) all equal 2 as x approaches 1. However,only h(x) has the same function value as its limit at x = 1.


Some functions have limits at every point.

If f is the identity function f (x) = x , then for any value of x0,

limx→x0

f (x) = limx→x0

x = x0.


Some functions have limits at every point.

If f is the constant function f (x) = k (function with the constant valuek), then for any value of x0,

limx→x0

f (x) = limx→x0

k = k.


Functions may fail to have a limit at a point in its domain.

The function U defined by

U(x) =

{0 if x < 0

1 if x ≥ 0

is the unit step function which hasno limit as x → 0 because itsvalues jump at x = 0.

For negative values of x arbitrarily close to zero, U(x) = 0.

For positive values of x arbitrarily close to zero, U(x) = 1.

There is no single value L approached by U(x) as x → 0.



The function g defined by

g(x) =

{1/x if x 6= 0

0 if x = 0

grows too large to have a limit.

g(x) has no limit as x → 0 because the values of g grow arbitrarilylarge in absolute value as x → 0 and do not stay close to any realnumber.



The function f defined by

f (x) =

{0 if x ≤ 0

sin 1x if x > 0

oscillates too much to have alimit.

f (x) has no limit as x → 0 because the function’s values oscillatebetween +1 and −1 in every open interval containing 0.

The values do not stay close to any one number as x → 0.


Using Calculators and Computers to Estimate Limits

Using a calculator or computer, one may guess a limit numerically as xgets closer and closer to x0. That procedure would be successful for thelimits of functions having equality of “limit” and “function value atthe point”.

However, calculators and computers can give false values andmisleading impressions for functions that are undefined at a point or failto have a limit there.

The differential calculus will help us to know when a calculator orcomputer is providing strange or ambiguous information about a function’sbehavior near some point.

For now, we simply need to be attentive to the fact that pitfalls may occurwhen using computing devices to guess the value of a limit.


Guessing a limit

Example 8.

Guess the value of

limx→0

√x2 + 100− 10

x2.


Solution


What is the answer?

The table lists values of the function for several values near x = 0.

As x approaches 0 through the values ±1,±0.5,±0.10, and ±0.01, thefunction seems to approach the number 0.05.

As we take ever smaller values of x , ±0.0005,±0.0001,±0.00001, and±0.000001, the function appears to approach the value 0.

So what is the answer? Is it 0.05 or 0, or some other value?

The calculator/computer values are ambiguous, but the theorems on limitswill confirm the correct limit value to be 0.05.

We shall now recall few theorems on limits.


Theorems on Limits

Using a calculator or computer, one may guess a limit numerically as xgets closer and closer to x0. That procedure would be successful for thelimits of functions having equality of “limit” and “function value atthe point”.

However, calculators and computers can give false values andmisleading impressions for functions that are undefined at a point or failto have a limit there. We discussed an example (in the last lecture) toconclude that the calculator/computer values are ambiguous.

In the lecture, we shall see some theorems for calculating limits. They tellhow to calculate limits of functions that are arithmetic combinations offunctions whose limits we already know.


Limit Laws : Sum and Difference Rules

Theorem 9 (Sum and Difference Rules).

Let L,M and c be real numbers and

limx→c

f (x) = L and limx→c

g(x) = M.

1. The limit of the sum of two functions is the sum of their limits.

limx→c

{f (x) + g(x)

}= L + M.

2. The limit of the difference of two functions is the difference oftheir limits.

limx→c

{f (x)− g(x)

}= L−M.


Limit Laws : Product and Constant Multiple Rules

Theorem 10 (Product and Constant Multiple Rules).

Let L,M, c and k be real numbers and

limx→c


g(x) = M.

1. The limit of the product of two functions is the product of theirlimits.

limx→c

{f (x)g(x)

}= LM.

2. The limit of a constant times a function is the constant timesthe limit of the function.

limx→c

{k f (x)

}= k L.


Limit Law : Quotient Rule

Theorem 11 (Quotient Rule).

Let L,M and c be real numbers and

limx→c


g(x) = M 6= 0.

The limit of a quotient of two functions is the quotient of theirlimits, provided the limit of the denominator is not zero.

limx→c

f (x)

g(x)=

L

M, M 6= 0.


Limit Law : Power Rule

Theorem 12 (Power Rule).

Let L and c be real numbers and

limx→c

f (x) = L.

If n is a positive integer, then

limx→c

(f (x)

)n= Ln.


Limit Law : Root Rule

Theorem 13 (Root Rule).


limx→c

f (x) = L.

If n is a positive integer, then

limx→c

n√

f (x) =n√

L.

If n is even, we assume that limx→c f (x) = L > 0.


Limit Law

Combining the above two theorems, we have the following.

Theorem 14.


limx→c

f (x) = L.

If r and s are integers with no common factor and s 6= 0, then

limx→c

(f (x)

)r/s= Lr/s

provided that Lr/s is a real number. (If s is even, we assume that L > 0.)

The limit of a rational power of a function is that power of the limitof the function, provided the latter is a real number.


Limit Law : Power Rule

Example 15.

Use the theorems on limits to find the following limits.

1. limx→c(x3 + 4x2 − 3)

2. limx→cx4+x2−1x2+5

3. limx→−2

√4x2 − 3


Solution

1. Apply the rules in the following order.

(a) Sum and Difference Rules(b) Power and Multiple Rules

Hence the limit is c3 + 4c2 − 3.


(a) Quotient Rule(b) Sum and Difference Rules(c) Power or Product Rule

Hence the limit is c4+c2−1c2+5

.


(a) Root Rule with n = 2(b) Difference Rules(c) Product and Multiple Rules

Hence the limit is√

13.


Limits of polynomials can be found by substitution.

To evaluate the limit of a polynomial function as x approaches c , merelysubstitute c for x in the formula for the function.

Theorem 16.

If P(x) = anxn + an−1xn−1 + · · ·+ a0, then

limx→c

P(x) = P(c) = ancn + an−1cn−1 + · · ·+ a0.


Limits of rational functions can be found by substitutionprovided the limit of the denominator is not zero.

To evaluate the limit of a rational function as x approaches a point c atwhich the denominator is not zero, substitute c for x in the formula forthe function.

Theorem 17.

If P(x) and Q(x) are polynomials and Q(c) 6= 0, then

limx→c

P(x)

Q(x)=

P(c)

Q(c).


Eliminating zero denominators algebraically

If the numerator P(x) and denominator Q(x) of a rational function P(x)Q(x)

of x are both zero at x = c , they have (x − c) as a common factor.

Canceling common factors in the numerator and denominator, theexpression may be reduced to the one whose denominator is no longer zeroat c .

If this happens, we can find the limit by substitution in the simplifiedfraction.

Example 18.

Evaluate

limx→1

x2 + x − 2

x2 − x.


Solution

Canceling the (x − 1)’s gives a simple fraction with the same values as theoriginal for x 6= 1:

x2 + x − 2

x2 − x=

x + 2

x, if x 6= 1.

Using the simple fraction, we find the limit of these values as x → 1 bysubstitution:

limx→1

x2 + x − 2

x2 − x=

x + 2

x=

1 + 2

1= 3.


Solution (contd..)

The graph of f (x) = x2+x−2x2−x is the same as the graph of g(x) = x+2

xexcept at x = 1, where f is undefined.

The functions have the same limit at x → 1.


Creating and canceling a common factor

Example 19.

Evaluate

limx→0

√x2 + 100− 10

x2.


Solution


Solution (contd...)

The table lists values of the function for several values near x = 0.

As x approaches 0 through the values ±1,±0.5,±0.10, and ±0.01, thefunction seems to approach the number 0.05.

As we take ever smaller values of x , ±0.0005,±0.0001,±0.00001, and±0.000001, the function appears to approach the value 0.

So what is the answer? Is it 0.05 or 0, or some other value?


Solution (contd...)

The calculator/computer values are ambiguous, but we can create acommon factor by multiplying both numerator and denominator by theconjugate radical expression

√x2 + 100 + 10 (obtained by changing the

sign after the square root). We get√

x2 + 100− 10

x2=

√x2 + 100− 10

x2.

√x2 + 100 + 10√x2 + 100 + 10

=1√

x2 + 100 + 10.

Therefore

limx→0

√x2 + 100− 10

x2=

1

20= 0.05.

This calculation provides the correct answer, in contrast to the ambiguouscomputer results discussed in the beginning of the solution.


Creating and canceling a common factor

Example 20.

Evaluate

limx→0

4−√

16 + x

x.


Solution

We cannot substitute x = 0 and the numerator and denominator have noobvious common factors.

However, we can create a common factor by multiplying both numeratorand denominator by the expression 4 +

√16 + x .

4−√

16 + x

x=

4−√

16 + x

x.4 +√

16 + x

4 +√

16 + x

=−1

4 +√

16 + x.

Therefore,

limx→0

4−√

16 + x

x= lim

x→0

−1

4 +√

16 + x.

=−1

8.


Sandwich Theorem

Sandwich Theorem refers to a function f whose values are sandwichedbetween the values of two other functions g and h that have the samelimit L at a point c .

Being trapped between the values of two functions that approach L,the values of f must also approach L.


Sandwich Theorem

Theorem 21.

Suppose that g(x) ≤ f (x) ≤ h(x) for all x in some open intervalcontaining c, except possibly at x = c itself. Suppose that

limx→c

g(x) = limx→c

h(x) = L.

Then limx→c f (x) = L.

The Sandwich Theorem is also called the Squeeze Theorem or thePinching Theorem.

The graph of f is sandwichedbetween the graphs of g and h.


Sandwich Theorem

We do not need the inequality to hold true for all values of x , exceptpossibly at x = c itself. We only need it to hold true on some open intervalcontaining c , except possibly at x = c itself (we call it, a puncturedneighborhood of c) so that we may apply the Sandwich Theorem.

The symbol, “≤” means that “less than”, or “equal to.” If we have

g(x) < f (x) < h(x)

or

g(x) ≤ f (x) ≤ h(x),

then as long as we have

limx→c

g(x) = limx→c

h(x) = L

we may apply the Sandwich Theorem.


Example

Example 22.

Given that

1− x2

4≤ u(x) ≤ 1 +

x2

2for all x 6= 0,

find limx→0 u(x), no matter how complicated u is.


Solution

Since

limx→0

(1− x2

4

)= 1 and lim

x→0

(1 +

x2

2

)= 1,

the Sandwich Theorem implies that limx→0 u(x) = 1.


Sandwich Theorem

Example 23.

The Sandwich Theorem helps us establish several important limit rules :

1. limθ→0 sin θ = 0

2. limθ→0 cos θ = 1

3. For any function f , limx→c |f (x)| = 0 implies that limx→c f (x) = 0.


Solution

(a) −|θ| ≤ sin θ ≤ |θ|, for all θ 6= 0. (b) 0 ≤ |θ| ≤ 1− cos θ, for all θ 6= 0.

(c) Since −|f (x)| ≤ f (x) ≤ |f (x)| and −|f (x)| and |f (x)| have limit 0 asx → c , it follows that

limx→0

f (x) = 0.


An important property

Theorem 24.

If

f (x) ≤ g(x)

for all x in some open interval containing c, except possibly at x = c itself,and the limits of f and g both exist as x approaches c, then

limx→c

f (x) ≤ limx→c

g(x).


An important property

The assertion resulting from replacing the less than or equal to (≤)inequality by the strict less than (<) inequality in the above Theorem isfalse.

The following figure shows that for θ 6= 0,

−|θ| < sin θ < |θ|,

but in the limit as θ → 0, equality holds.


Exercises

Exercises 25.

For the function f (t) graphed here, find the following limits or explain whythey do not exist.

(a) limt→−2 f (t)

(b) limt→−1 f (t)

(c) limt→0 f (t)

(d) limt→0.5 f (t).


Solutions

(a) 0

(b) −1

(c) Does not exist.As t approaches 0, from the left, f (t) approaches −1.As t approaches 0, from the right, f (t) approaches 1.

(d) 1.


Exercises

Exercises 26.

Which of the following statements about the function y = f (x) graphedhere are true, and which are false?

1. limx→2 f (x) does not exist.

2. limx→2 f (x) = 2.

3. limx→1 f (x) does not exist.

4. limx→x0 f (x) exists at every point x0 in (−1, 1).

5. limx→x0 f (x) exists at every point x0 in (1, 3).


Solutions

1. False

2. False

3. True

4. True

5. True


Exercises

Exercises 27.

Explain why the following limits do not exist.

1. limx→0x|x |

2. limx→11

x−1


Solutions

1. limx→0x|x | does not exist because x

|x | = 1 if x > 0 andx|x | = −1 if x < 0.

As x approaches 1, from the left, x|x | approaches −1.

As x approaches 1, from the right, x|x | approaches 1.

2. As x approaches 1, from the left, the values of 1x−1 become

increasingly large and negative.As x approaches 1, from the right, the values of 1

x−1 becomeincreasingly large and positive.


Exercises

Exercise 28.

Suppose that a function f (x) is defined for all x in [−1, 1]. Can anythingbe said about the existence of

limx→0

f (x) ?

Give reasons for your answer.


Solution

Nothing can be said.

In order for

limx→0

f (x)

to exist, f (x) must close to a single value for x near 0 regardless of thevalue f (0) itself.


Exercises

Exercise 29.

If f (1) = 5, must limx→1 f (x) exist? If it does, then must

limx→1

f (x) = 5 ?

Can we conclude anything about

limx→1

f (x) ?

Explain.


Solution

Nothing can be said because the existence of a limit depends on the valueof f (x) when x is near 1, not on f (1) itself.

If limx→1 f (x) exists, its value may be some number other than f (1) = 5.

We can conclude nothing about f (1) from

limx→1

f (x) = 5.


Exercises

Exercises 30.

Find the following :

1. limx→−7(2x + 5)

2. limh→03√

3h+1+1

3. limz→0(2z − 8)1/3

4. lims→2/3 3s(2s − 1)

5. limx→1

1x−1

x−1

6. limx→0

1x−1

+ 1x+1

x

7. limx→1x−1√x+3−2

8. limx→44−x

5−√x2+9

9. limx→0 sin2 x

10. limx→−π√

x + 4 cos(x + π)

11. limx→0(x2 − 1)(2− cos x)


Solutions

1. −9

2. 3/2

3. −2

4. 2/3

5. −1

6. −2

7. 4

8. 5/4

9. 0

10.√

4− π11. −1


Exercises

Exercise 31.

If √5− 2x2 ≤ f (x) ≤

√5− x2

for −1 ≤ x ≤ 1, find

limx→0

f (x).


Solution

limx→0

√5− 2x2 =

√5 and limx→0

√5− x2 =

√5.

By Sandwich Theorem,

limx→0

f (x) =√

5.


Exercises

Exercise 32.

It can be shown that the inequalities

1− x2

6<

x sin x

2− 2 cos x< 1

hold for all values of x close to zero. What, if anything, does this tell youabout

limx→0

x sin x

2− 2 cos x?

Give reasons for your answer.


Solution

Given that the inequalities

1− x2

6<

x sin x

2− 2 cos x< 1

hold for all values of x close to zero.

We can find an open interval containing 0, except possibly at x = 0 itselfin which the above inequalities hold true.

Also, limx→0

(1− x2

6

)= 1 and limx→0 1 = 1.


limx→0

x sin x

2− 2 cos x= 1.


Exercises

Exercise 33.

If

limx→4

f (x)− 5

x − 2= 1,

find

limx→4

f (x).


Solution

1 = limx→4

f (x)− 5

x − 2

=limx→4 f (x)− limx→4 5

limx→4 x − limx→4 2

Hence

limx→4

f (x) = 7.


Exercises

Exercise 34.

If

limx→0

f (x)

x2= 1,

find

1. limx→0 f (x)

2. limx→0f (x)x .


Solution

1. limx→0 f (x) = limx→0f (x)x2 limx→0 x2 = 1.0 = 0

2. limx→0f (x)x = limx→0

f (x)x2 limx→0 x = 1.0 = 0


Exercises

Exercise 35.

1. Graph

g(x) = x sin(1/x)

to estimate

limx→0

g(x),

zooming in on the origin as necessary.

2. Confirm your estimate in part (a) with a proof.


Solution

1. limx→0 x sin(1/x) = 0

2. −1 ≤ sin 1x ≤ 1 for x 6= 0.

Since x > 0 =⇒ −x ≤ x sin 1x ≤ x , by Sandwich Theorem,

limx→0

x sin(1/x) = 0.

Since x < 0 =⇒ −x ≥ x sin 1x ≥ x , by Sandwich Theorem,

limx→0

x sin(1/x) = 0.


Exercises

Exercise 36.

1. Graph

h(x) = x2 cos(1/x3)

to estimate

limx→0

h(x),

zooming in on the origin as necessary.

2. Confirm your estimate in part (a) with a proof.


Solution

1. limx→0 x2 cos(1/x3) = 0

2. Since −1 ≤ cos 1x3 ≤ 1 for x 6= 0, −x2 ≤ x2 cos(1/x3) ≤ x2.


limx→0

x2 cos(1/x3) = 0.


Definition of a limit

We now recall the following “informal definition” of a limit.

If f (x) gets arbitrarily close to L (as close to L as we like) for all xsufficiently close to x0, we say that f approaches the limit L as xapproaches x0, we write

limx→x0

f (x) = L.

This definition is “informal” because phrases like arbitrarily close andsufficiently close are imprecise ; their meaning depends on the context.


Limit of a function

Definition 37.

Let f (x) be defined on an open interval about x0, except possibly at x0

itself. We say that the limit of f (x) as x approaches x0 is the numberL, and write

limx→x0

f (x) = L

if, for every number ε > 0, there exists a corresponding number δ > 0 suchthat for all x,

0 < |x − x0| < δ =⇒ |f (x)− L| < ε.

The formal definition of limit does not tell how to find the limit of afunction, but its enables us to verify that a suspected limit iscorrect.


Limit of a function


For a given ε (challenge), we find δ (response)

Challenge : For a given ε, weshould make

|f (x)− L| < ε.

Response : We should find δ, suchthat whenever x satisfying

0 < |x − x0| < δ,

f (x) has to satisfy

|f (x)− L| < ε.


Example : Testing Definition

The following example shows how the definition can be used to verify limitstatement for specific functions.












If ε is not given to be any specific number, then what is δ?

We may find algebraically a δ for a given f , L, x0 and ε > 0.


How to find algebraically a δ for a given f , L, x0 and ε > 0

The process of finding a δ > 0 such that for all x

0 < |x − x0| < δ =⇒ |f (x)− L| < ε

can be accomplished in two steps.

1. Step : 1 Solve the inequality |f (x)−L| < ε to find an open interval(a, b) containing x0 on which the inequality holds for all x 6= x0.

2. Step : 2 Find a value of δ > 0 that places the open interval(x0 − δ, x0 + δ) centered at x0 inside the interval (a, b). Theinequality |f (x)− L| < ε will hold for all x 6= x0 in this δ-interval.


Different Notations : Neighborhoods

For a fixed real number L and a positive real number ε, we call the set{y : |y − L| < ε

}an ε-neighborhood of L, denoted by N(L, ε).

Whereas the set {y : 0 < |y − L| < ε

}is called a deleted (punctured) ε-neighborhood of L, denoted byNd(L, ε). Note that Nd(L, ε) = N(L, ε) \ {L}.Let A be a subset of R (the set of all real numbers). A point x0 is said tobe an accumulation point of a set A if any neighborhood of x0 containsan element of A distinct from x0.


Rephrasing the ε− δ definition of a limit in terms ofneighborhoods.

Definition 38.

Let f : D → R, where D ⊆ R, and suppose that x0 ∈ R is anaccumulation point of D. Then

limx→x0

f (x) = L

if and only if for every neighborhood V of L, there is a puncturedneighborhood U of x0 such that

x ∈ D ∩ U implies that f (x) ∈ V .

There is an equivalent “sequential characterization” of the limit, whichwill be discussed later.


Example

Example 39.

Show that

limx→1

(5x − 3) = 2.


Solution

Let ε > 0 be given.

For x 6= 1, we have f (x) = 5x − 3, and we need to solve the inequality

|(5x − 3)− 2| < ε. (1)

The inequality (1) becomes

|x − 1| < ε/5.

Thus, we can δ to be ε/5, or, any positive number less that ε/5. Here, wetake δ = ε/5.

If 0 < |x − 1| < δ, then |(5x − 3)− 2| = |5x − 5| = 5|x − 1| < 5(ε/5) = ε.


Example : Limit of the identity function

Example 40.

Prove that

limx→x0

x = x0.


Solution

Let ε > 0 be given. We must find δ > 0 such that for all x

0 < |x − x0| < δ implies |x − x0| < ε.

The implication will hold if δ equals ε or any positive number smaller thanε.

This proves that

limx→x0

x = x0.


Example : Limit of the constant function

Example 41.

Prove that

limx→x0

k = k (any constant).


Solution

Let ε > 0 be given. We must find δ > 0 such that for all x

0 < |x − x0| < δ implies |k − k | < ε.

Since k − k = 0, we can use any positive number for δ and implication willhold.

This proves that

limx→x0

k = k.


Example

Example 42.

For the limit

limx→5

√x − 1 = 2

find a δ > 0 that works for ε = 1.

That is, find a δ > 0 such that for all x

0 < |x − 5| < δ =⇒ |√

x − 1− 2| < 1.


Solution

Step 1 We first solve the inequality |√

x − 1− 2| < 1 to find an intervalcontaining 5 on which the inequality holds for all x 6= 5.

|√

x − 1− 2| < 1

−1 <√

x − 1− 2 < 1

2 < x < 10.

The inequality holds for all x in the open interval (2, 10), so it holds for allx 6= 5 in this interval as well.


Solution (contd...)

Step 2

Find a value of δ > 0 to place the centered interval 5− δ < x < 5 + δ(centered at x0 = 5) inside the interval (2, 10). The distance from 5 to thenearer endpoint of (2, 10) is 3.

If we take δ = 3, or any positive number less than 3, then for all x

0 < |x − 5| < 3 implies |√

x − 1− 2| < 1.


Example

Example 43.

If

f (x) =

{x2 if x 6= 2

1 if x = 2,

prove that

limx→2

f (x) = 4.


Solution


Step 1 We first solve the inequality |x2 − 4| < ε to find an intervalcontaining 2 on which the inequality holds for all x 6= 2.

|x2 − 4| < ε

4− ε < x2 < 4 + ε

As ε is a positive number, we now discuss the case when ε < 4.

Case : ε < 4|x2 − 4| < ε√

4− ε < |x | <√

4 + ε√4− ε < x <

√4 + ε.

The inequality holds for all x 6= 2 in the open interval (√

4− ε,√

4 + ε).


Solution (contd...)

Step 2

We now find a δ > 0 that places the centered interval (2− δ, 2 + δ) insidethe interval

(√

4− ε,√

4 + ε).

Take δ to be the distance from x0 = 2 to the nearer endpoint of(√

4− ε,√

4 + ε).

In other words, take δ = min{

2−√

4− ε,√

4 + ε− 2}

, or, any positive

number smaller than this minimum. For all x ,

0 < |x − 2| < δ implies |f (x)− 4| < ε.

This completes the proof for ε < 4.


Solution (contd...)

Case : ε ≥ 4|x2 − 4| < ε

4− ε < x2 < 4 + ε

0 < x2 < 4 + ε

0 < x <√

4 + ε.

Take δ to be the distance from x0 = 2 to the nearer endpoint of(0,√

4 + ε).

In other words, take δ = min{

2,√

4 + ε− 2}

, or, any positive number

smaller than this minimum. For all x ,

0 < |x − 2| < δ implies |f (x)− 4| < ε.

This completes the proof.


Using the definition to prove theorems

We do not usually rely on the formal definition of limit to verify specificlimits. Rather we appeal to general theorems about limits.

The definition is used to prove these theorems.

Example 44 (Proving the rule for the limit of a sum).

Given that limx→c f (x) = L and limx→c g(x) = M, prove that

limx→c

{f (x) + g(x)

}= L + M.


Solution


We want to find a positive number δ such that for all x ,

0 < |x − c | < δ implies |f (x) + g(x)− (L + M)| < ε.

We have

|f (x) + g(x)− (L + M)| ≤ |f (x)− L|+ |g(x)−M|.

From the previous inequality, to get

|f (x) + g(x)− (L + M)| < ε

it is observed that we can apply the definition of limit for ε/2 to the limitsL,M for the functions f (x), g(x) respectively.


Solution (contd...)

Since limx→c f (x) = L, there exists a positive number δ1 such that for allx ,

0 < |x − c| < δ1 implies |f (x)− L| < ε/2.

Similarly, since limx→c g(x) = M, there exists a positive number δ2 suchthat for all x ,

0 < |x − c | < δ2 implies |g(x)−M| < ε/2.


Solution (contd...)

Let δ = min{δ1, δ2

}, or any positive number smaller than the minimum.

If 0 < |x − c | < δ, then

|x − c | < δ1, so |f (x)− L| < ε/2,

and

|x − c| < δ2, so |g(x)−M| < ε/2.

Therefore

|f (x) + g(x)− (L + M)| ≤ ε/2 + ε/2 = ε.

This shows that

limx→c

{f (x) + g(x)

}= L + M.


Showing L is not a limit.

What is a number L not the limit of f (x) as x → x0?

We can prove that limx→x0 f (x) 6= L by providing an ε > 0 such that nopossible δ > 0 satisfies the condition

for all x , 0 < |x − x0| < δ =⇒ |f (x)− L| < ε.


Showing L is not a limit.

We accomplish this for our candidate ε by showing that for each eachδ > 0 there exists a value of x such that

0 < |x − x0| < δ and |f (x)− L| ≥ ε.


Exercises

Exercise 45.

Sketch the interval (a, b) on the x-axis with the point x0 inside. Then finda value of δ > 0 such that for all x , 0 < |x − x0| < δ =⇒ a < x < b.

1. a = 1, b = 7, x0 = 5

2. a = −7/2, b = −1/2, x0 = −3

3. a = 2.7591, b = 3.2391, x0 = 3


Solution

1. |x − 5| < δ =⇒ −δ + 5 < x < δ + 5.Since a = 1, −δ + 5 = 1 gives δ = 4. Since b = 7, δ + 5 = 7 givesδ = 2. The value of δ which assures |x − 5| < δ =⇒ 1 < x < 7 isthe min{4, 2} = 2.

2. The value of δ which assures |x − (−3)| < δ =⇒ −72 < x < −1

2 isthe min{5

2 ,12} = 1

2 .

3. The value of δ which assures |x − 3| < δ =⇒ 2.7591 < x < 3.2391is the min{0.2391, 2.7591} = 0.2391.


Exercises

Exercise 46.

Use the graph to find a δ > 0 such that for all x

0 < |x − 1| < δ =⇒ |f (x)− 1| < ε.


Solution

From the graph, the value of δ which assures

|x − 1| < δ =⇒ |f (x)− 1| < ε

is the

min{ 9

16,

7

16

}=

7

16.


Exercises

Exercise 47.


0 < |x − 1| < δ =⇒ |f (x)− 3| < ε.


Solution


|x − 1| < δ =⇒ |f (x)− 3| < ε

is the

min{2−

√3

2,

√5− 2

2

}=

√5− 2

2.


Exercises

Exercise 48.


0 < |x − x0| < δ =⇒ |f (x)− L| < ε.


Solution


|x − 1/2| < δ =⇒ |f (x)− 2| < ε

is the

min{ 1

1.99− 1

2,

1

2− 1

2.01

}=

1

2− 1

2.01.


Exercises

Exercise 49.

Each of the following exercises gives a function f (x) and numbers L, x0,and ε > 0. In each case, find an open interval about x0 on which theinequality |f (x)− L| < ε holds. Then give a value for δ > 0 such that forall x satisfying 0 < |x − x0| < δ the inequality |f (x)− L| < ε holds.

1. f (x) = x + 1, L = 5, x0 = 4, ε = 0.01

2. f (x) =√

x − 7, L = 4, x0 = 23, ε = 1

3. f (x) = 1/x , L = −1, x0 = −1, ε = 0.1

4. f (x) = mx + b, m > 0, L = (m/2) + b, x0 = 1/2, ε = c > 0


Solution

1. δ = 0.01

2. δ = 7

3. δ = 111

4. δ = cm


Exercises

Exercise 50.

Each of the following exercises gives a function f (x), a point x0, and apositive number ε. Find L = limx→x0 f (x). Then find a number δ > 0 suchthat for all x

0 < |x − x0| < δ =⇒ |f (x)− L| < ε.

1. f (x) = 3− 2x , x0 = 3, ε = 0.02

2. f (x) = x2+6x+5x+5 , x0 = −5, ε = 0.05

3. f (x) = 4/x , x0 = 2, ε = 0.4


Solution

1. δ = 0.01

2. δ = 0.05

3. δ = 13


Exercises

Exercise 51.

Prove the limit statements in the following statement.

1. limx→4(9− x) = 5

2. limx→−2 f (x) = 4 if

f (x) =

{x2 ifx 6= −2

1 ifx = −2

3. limx→0 f (x) = 0 if

f (x) =

{2x ifx < 0

x/2 ifx ≥ 0


Solution

1. δ = min{4ε− ε2, 4ε+ ε2} = 4ε− ε2

2. δ = min{√

4 + ε− 2, 2−√

4− ε}3. δ = ε

2


Exercises

Exercise 52.

Prove the following limit statement.

limx→0

x2 sin1

x= 0.


Solution

By the figure,

−x2 ≤ x2 sin1

x≤ x2

for all x except possibly at x = 0.

Since

limx→0

(−x2) = 0 = limx→0

x2,

then by the Sandwich Theorem,

limx→0

x2 sin1

x= 0.


Exercises

Exercise 53.

Prove that

limx→c

f (x) = L ⇐⇒ limh→0

f (h + c) = L.


Solution

Write x = h + c .

Then

0 < |x − c | < δ =⇒ 0 < |h − 0| < δ.

The following are equivalent :

1. limx→c f (x) = L

2. for any ε > 0, there exists δ > 0 such that |f (x)− L| < ε whenever0 < |x − c | < δ.

3. for any ε > 0, there exists δ > 0 such that |f (h + c)− L| < εwhenever 0 < |h − 0| < δ.

4. limh→0 f (h + c) = L.


One-sided limits

We now extend the limit concept to one-sided limits, which are limits asx approaches the number c

from the left-hand side (where x < c)

or

the right-hand side (x > c) only.


One-sided limits

To have a limit L as x approaches c, a function f must be defined onboth sides of c and its values f (x) must approach L as x approaches cfrom either side. Because of this, ordinary limits are called two-sided.

If f fails to have a two-sided limit as c , it may still have a one-sided limit,that is, a limit if the approach is only from one side.

If the approach is from the right, the limit is a right-hand limit.

From the left, it is a left-hand limit.


One-sided limits

The function f (x) = x/|x | has limit 1 as x approaches 0 from the right,and limit −1 as x approaches 0 from the left.

Since these one-sided limit values are not the same, there is no singlenumber that f (x) approaches 0. So f (x) does not have a (two-sided) limitat 0.


Right-hand limit

If f (x) is defined on an interval (c , b), where c < b, and approachesarbitrarily close to L as x approaches c from within that interval, then fhas right-hand limit L at c .

We write

limx→c+

f (x) = L.

The symbol “x → c+” means thatwe consider only values of x greaterthan c .


Left-hand limit

If f (x) is defined on an interval (a, c), where a < c , and approachesarbitrarily close to M as x approaches c from within that interval, then fhas left-hand limit M at c .

We write

limx→c−

f (x) = M.

The symbol “x → c−” means thatwe consider only values of x lessthan c .


Example

Example 54.

The domain of f (x) =√

4− x2 in [−2, 2]. We have

limx→2+

√4− x2 = 0 and lim

x→2−

√4− x2 = 0.

The function does not have aleft-hand limit at x = −2 or aright-hand limit at x = 2. It doesnot have ordinary two-sided limitsat either −2 or 2.


Properties of one-sided limits

One-sided limits have all the properties dicussed in the last lecture.

The right-hand limit of the sum of two functions is the sum of theirright-hand limits, and so on.

The theorems for limits of polynomials and rational functions hold withone-sided limits, as does the Sandwich Theorem.


One-sided limits are related to limits

One-sided limits are related to limits in the following way.

Theorem 55.

A function f (x) has a limit as x approaches c if and only if it has left-handand right hand limits there and these one-sided limits are equal :

limx→c

f (x) = L ⇐⇒ limx→c−

f (x) = L and limx→c+

f (x) = L.


Exercises

Exercise 56.

If

x4 ≤ f (x) ≤ x2

for x in [−1, 1] and

x2 ≤ f (x) ≤ x4

for x < −1 and x > 1, at what points c do you automatically know

limx→c

f (x) ?

What can you say about the value of the limit at these points?


Solution

limx→c f (x) exists at those points c where

limx→c

x4 = limx→c

x2.

Thus, c4 = c2 =⇒ c = 0, 1, or −1. Moreover,

limx→0

f (x) = 0

and

limx→−1

f (x) = limx→1

f (x) = 1.


Precise definition of one-sided limits

We say that f (x) has right-hand limit L at x0, and write

limx→x+

0

f (x) = L

if for every number ε > 0 there exists a corresponding number δ > 0 suchthat for all x

x0 < x < x0 + δ =⇒ |f (x)− L| < ε.


Precise definition of one-sided limits

We say that f (x) has left-hand limit L at x0, and write

limx→x−0

f (x) = L

if for every number ε > 0 there exists a corresponding number δ > 0 suchthat for all x

x0 − δ < x < x0 =⇒ |f (x)− L| < ε.


Using ε− δ definition, proving limit

Example 57.

Using ε− δ definition, prove that

limx→0+

√x = 0.


Proof

Start with : Let ε > 0 be given.

We want to find a δ > 0 such that for all x

0 < x < δ =⇒ |√

x − 0| < ε.

Squaring both sides of the this last inequality gives

x < ε2 if 0 < x < δ.

If we choose δ = ε2, we have

0 < x < δ < ε2 =⇒√

x < ε.

This shows that

limx→0+

√x = 0.


Example

The functions examined so far have had some kind of limit at each pointof interest.

In general, that need not be the case

Example 58.

Show that y = sin(1/x) has no limit as x approaches zero from either side.


Solution

As x approaches zero, its reciprocal, 1/x , grows without bound and thevalues of sin(1/x) cycle repeatedly from −1 to 1.

There is no single number L that the function’s values stay increasinglyclose to as x approaches zero.

This is true even if we restrict x to positive values or to negative values.

The function has neither a right-hand limit nor a left-hand limit at x = 0.


Limits involving sin θ/θ

A central fact about (sin θ/θ) is that in radian measure its limit as θ →is 1.

We can see this in the following figure and confirm it algebraically usingthe Sandwich Theorem.


Limits involving sin θ/θ

Theorem 59.

If θ is in radians, then sin θθ = 1.

The plan is to show that right-hand and left-hand limits are both 1.

Then we will know that the two-sided limit is 1 as well.

Proof is omitted.


Example

Example 60.

Show that

(a) limh→0cos h−1

h = 0 and

(b) limx→0sin 2x

5x = 25 .


Solution

(a) We have the half-angle formula

cos h = 1− 2 sin2(h/2).

limh→0

cos h − 1

h= lim

h→0

−2 sin2(h/2)

h= −(1)(0) taking θ = h/2

(b) We need a 2x in the denominator, not a 5x . We produce it bymultiplying numerator and denominator by 2/5.

limx→0

sin 2x

5x= lim

x→0

(2/5). sin 2x

(2/5).5x

=2

5(1) taking θ = 2x


Example

Example 61.

Find

limt→0

tan t sec 2t

3t.


Solution

From the definition of tan t and sec 2t, we have

limt→0

tan t sec 2t

3t= lim

t→0

sin t

t.

1

cos t.

1

cos 2t

= =1

3(1)(1)(1) =

1

3.


Exercises

Exercise 62.

Which of the following statements about the function y = f (x) graphedhere are true, and which are false?page 71 graph exer 1

(a) limx→1+ f (x) = 1

(b) limx→0− f (x) = 0

(c) limx→0− f (x) = 1

(d) limx→0− f (x) = limx→0+ f (x)

(e) limx→0 f (x) exists.

(f) limx→0 f (x) = 0

(g) limx→0 f (x) = 1

(h) limx→1 f (x) = 1

(i) limx→1 f (x) = 0

(j) limx→2− f (x) = 2

(k) limx→1− f (x) does not exist.

(l) limx→2+ f (x) = 0


Solution

(a) True

(b) True

(c) False

(d) True

(e) True

(f) True

(g) False

(h) False

(i) False

(j) False

(k) True

(l) False


Exercises

Exercise 63.

Let

f (x) =

{3− x if x < 2x2 + 1 if x > 2.

(a) Find limx→2+ f (x) and limx→2− f (x).

(b) Does limx→2 f (x) exist ? If so, what is it ? If not, why not ?

(c) Find limx→4− f (x) and limx→4+ f (x) .

(d) Does limx→4 f (x) exist ? If so, what is it ? If not, why not ?P. Sam Johnson Introduction August 7, 2019 170/288

Solution

(a) limx→2+ f (x) = 22 + 1 = 2 and limx→2− f (x) = 3− 2 = 1.

(b) No. limx→2 f (x) does not exist because limx→2+ f (x) 6= limx→2− f (x).

(c) limx→4− f (x) = 42 + 1 = 3 and limx→4+ f (x) = 4

2 + 1 = 3.

(d) Yes. limx→4 f (x) = 3 because 4 = limx→4− f (x) = limx→4− f (x).


Exercises

Exercise 64.

Let

f (x) =

{0 if x ≤ 0

sin 1x if x > 0.

(a) Does limx→0+ f (x) exist ? If so, what is it ? If not, why not ?

(b) Does limx→0− f (x) exist ? If so, what is it ? If not, why not ?

(c) Does limx→0 f (x) exist ? If so, what is it ? If not, why not ?


Solution

(a) No. limx→0+ f (x) does no exist since sin( 1x ) does not approach any

single value as x approaches 0.

(b) limx→0− f (x) = 0

(c) limx→0 f (x) does not exist because limx→0+ f (x) does not exist.


Exercises

Exercise 65.

Let g(x) =√

x sin(1/x).

(a) Does limx→0+ g(x) exist ? If so, what is it ? If not, why not ?

(b) Does limx→0− g(x) exist ? If so, what is it ? If not, why not ?

(c) Does limx→0 g(x) exist ? If so, what is it ? If not, why not ?


Solution

(a) Yes. limx→0+ g(x) = 0 by the Sandwich Theorem since−√

x ≤ g(x) ≤√

x when x > 0.

(b) No. limx→0− g(x) does not exist since√

x is not defined for x < 0.

(c) No. limx→0 g(x) does not exist since limx→0− g(x) does not exist.


Exercises

Exercise 66.

(a) Graph

f (x) =

{1− x2 if x 6= 1

2 if x = 1.

(b) Find limx→1+ f (x) and limx→1− f (x).

(c) Does limx→1 f (x) exist ? If so, what is it ? If not, why not ?


Solution

(a)

(b) limx→1+ f (x) = 0 = limx→1− f (x).

(c) Yes. limx→1 f (x) = 0 since the right-hand and left-hand limits existand equal 0.


Exercises

Exercise 67.

Graph the function given below. Then answer these questions.

(a) What are the domain and range of f ?

(b) At what points c, if any, does lim x → cf (x) exist ?

(c) At what points does only the left-hand limit exist ?

(d) At what points does only the right-hand limit exist ?

f (x) =

x if − 1 ≤ x < 0, or 0 < x ≤ 1

1 if x = 0

0 if x < −1, or x > 1


Solution

(a) domain : −∞ < x <∞ and rang : −1 ≤ y ≤ 1

(b) lim x → cf (x) exists for c belonging to (∞,−1) ∪ (−1, 1) ∪ (1,∞)

(c) none

(d) none


Exercises

Exercises 68.

Find the limits in the following exercises.

1. limx→1+

√x−1x+2

2. limx→1−

(1

x+1

)(x+6x

)(3−x

7

)3. limx→+

√x−1x+2

4. (a) limx→2+ (x + 3) |x+2|x+2

(b) limx→2−(x + 3) |x+2|x+2


Solution

1. 0

2. 1

3. (a) 1(b) −1


Exercises

Exercise 69.

The function whose value at any number x is the greatest integer less thanor equal to x is called the greatest integer function or the integer floorfunction. It is denoted bxc.

The graph of the greatest integerfunction y = bxc lies on or bellowthe line y = x, so it provides aninteger floor for x.Use the graph of the greatestinteger function y = |x |, find thelimits in the following exercises.

1. (a) limθ→3+bθcθ

(b) limθ→3−bθcθ


Solution

1. (a) limθ→3+bθcθ = 1

(b) limθ→3−bθcθ = 2

3


Exercises

Exercise 70.

Using limθ→0sin θθ = 1, find the limits in the following exercises.

1. limθ→0sin√

2θ√2θ

2. limh→0−h

sin 3h

3. limx→0x csc 2xcos 5x

4. limθ→01−cos θ

sin 2θ

5. limh→0sin(sin h)

sin h

6. limθ→0 cos θ

7. limy→0sin 3y cot 5yy cot 4y

8. limθ→0tan θ

θ2 cot 3θ


Solution

1. limθ→0sin√

2θ√2θ

= 1

2. limh→0−h

sin 3h = 13

3. limx→0x csc 2xcos 5x

12

4. limθ→01−cos θ

sin 2θ = 0

5. limh→0sin(sin h)

sin h = 1

6. limθ→0 cos θ = 0

7. limy→0sin 3y cot 5yy cot 4y = 12

5

8. limθ→0tan θ

θ2 cot 3θ= 3


Exercises

Exercise 71.

Once you know

limx→a+

f (x)

and

limx→a−

f (x)

at an interior point of the domain of f , do you then know limx→a f (x) ?Give reasons for your answer.


Solution

Yes.

If

limx→a+

f (x) = L = limx→a−

f (x),

then limx→a f (x) = L.

If

limx→a+

f (x) 6= limx→a−

f (x),

then limx→a f (x) does not exist.


Exercises

Exercise 72.

If you know that

limx→c

f (x)

exists, can you find its value by calculating limx→c+ f (x) ?Give reasons for your answer.


Solution

Since

limx→c

f (x) = L ⇐⇒ limx→c−

f (x) = L and limx→c+

f (x) = L,

then limx→c f (x) can be found by calculating limx→c+ f (x).


Exercises

Exercise 73.

Suppose that f is an even function of x. Does knowing that

limx→2−

f (x) = 7

tell you anything about either limx→2− f (x) or limx→2+ f (x) ?Give reasons for your answer.


Solution

If f is an even function of x , then f (−x) = f (x). Given

limx→2−

f (x) = 7,

then limx→2− f (x) = 7.

However, nothing can be said about

limx→2−

f (x)

because we don’t know limx→2+ f (x).


Exercises

Exercise 74.

Given ε > 0, find an interval I = (4− δ, 4), δ > 0, such that if x lies in I ,then

√4− x < ε. What limit is being verified and what is its value?


Solution

I = (4− δ, 4) =⇒ 4− δ < x < 4.

Also,√

4− x < ε =⇒ 4− x < ε2 =⇒ x < 5 + ε2.

Choose δ = ε2 =⇒ limx→4−√

4− x = 0.


Exercises

Exercise 75.

Use the definitions of right-hand and left-hand limits to prove thefollowing limit statement.

limx→2+

x − 2

|x − 2|= 1.


Solution

Since x → 2+, we have x > 2 and |x − 2| = x − 2.

Then∣∣∣ x−2|x−2| − 1

∣∣∣ < ε =⇒ 0 < ε which is always true so long as x > 2.

Hence we can choose any δ > 0, and thus

2 < x < 2 + δ =⇒∣∣∣ x−2|x−2| − 1

∣∣∣ < ε.

Thus limx→2+x−2|x−2| = 1.


Exercises

Exercise 76.

(a) limx→400+bxc and

(b) limx→400−bxc ; then use limite of definitions to verify your findings.

(c) Bases on your conclusions in parts (a) and (b), can you say anythingabout limx→400bxc ? Give reasons for your answer.


Solution

(a) limx→400+bxc = 400. Just observe that if 400 < x < 401, thenbxc = 400. Thus if we choose δ = 1, we have for any number ε > 0such that 400 < x < 400 + δ =⇒ |bxc − 400| = |400− 400| = 0 < ε.

(b) limx→400−bxc = 399. Just observe that if 399 < x < 400, thenbxc = 399. Thus if we choose δ = 1, we have for any number ε > 0such that 400− δ < x < 400 =⇒ |bxc − 399| = |399− 399| = 0 < ε.

(c) Since limx→400+bxc 6= limx→400−bxc, we conclude limx→400bxc doesnot exist.


Overview

When we plot function values generated in a laboratory or collected in thefield, we often connect the plotted points with an unbroken curve to showwhat the function’s values are likely to have been at the times we did notmeasure.


Overview

When we connect the plotted points with an unbroken curve, we areassuming that we are working with a continuous function, so its outputsvary continuously with the inputs and do not jump from one value toanother without taking on the values in between.

The limit of a continuous function as x approaches c can be found simplyby calculating the value of the function at c .

An function y = f (x) whose graph can be sketched over its domain in onecontinuous motion without lifting the pencil is an example of a continuousfunction.

In this lecture we investigate more precisely what it means for a functionto be continuous. We also study the properties of continuous functions.


Continuity at a Point

We start with an example.

Example 77 (Investigating Continuity).

Find the points at which the function f in the following figure iscontinuous and the points at which f is discontinuous.

The function is continuous on [0, 4]except at x = 1, x = 2, and x = 4.


Solution

Points at which f is continuous :At x = 0, limx→0+ f (x) = f (0).At x = 3, limx→3 f (x) = f (3).At 0 < c < 4, c 6= 1, 2, limx→c f (x) = f (c).

Points at which f is discontinuous :At x = 1, limx→1 f (x) does not exist.At x = 2, limx→2 f (x) = 1, but 1 6= f (2).At x = 4, limx→4− f (x) = 1, but 1 6= f (4).At c < 0, c ≥ 4, these points are not in the domain of f .


Continuity at a point in a function’s domain

Continuity at a point in a function’s domain, we need to define continuityat an interior point (which involves a two-sided limit) and continuity at anendpoint (which involves a one-sided limit).


Continuity at a point in a function’s domain

Definition 78.

Interior PointA function y = f (x) is continuous at an interior point c of its domain if

limx→c

f (x) = f (c).

EndpointA function y = f (x) is continuous at a left point a or is continuous ata right endpoint b of its domain if

limx→a+

f (x) = f (a) or limx→b−

f (x) = f (b), respectively.


Point of discontinuity

If a function f is not continuous at a point c, we say that f isdiscontinuous at c and c is a point of discontinuity of f .

Note c need not be in the domain of f .


Continuous from the right / left

A function f is right-continuous (continuous from the right) at a pointx = c in its domain if

limx→c+

f (x) = f (c).

It is left-continuous (continuous from the left) at c if

limx→c−

f (x) = f (c).

Thus, a function is continuous at a left endpoint a of its domain if it isright-continuous at a ; continuous at a right endpoint b of its domain if itis left-continuous at b.



A function is continuous at an interior point c of its domain if and only ifit is both right-continuous and left-continuous at c .


The epsilon-delta definition of continuity at x = x0

Definition 79.

Let f (x) be defined on an open interval about x0, including at x0 itself.We say that the function f is continuous at x = x0 if, for every numberε > 0, there exists a corresponding number δ > 0 such that for all x,

|x − x0| < δ =⇒ |f (x)− f (x0)| < ε.



The unit step function U defined by

U(x) =

{0 if x < 0

1 if x ≥ 0

is right-continuous at c = 0, but not left-continuous.



The greatest integer function

y = bxc

is continuous at every noninteger point.

It is right-continuous, but not left-continuous, at every integer point.


Continuity Test

We summarize continuity at a point in the form of a test.

A function f (x) is continuous at x = c if and only if it meets the followingthree conditions.

1. f (c) exists (c lies in the domain of f )

2. limx→c f (x) exists (f has a limit as x → c)

3. limx→c f (x) = f (c) (the limit equals the function value)

For one-sided continuity and continuity at an endpoint, the limits of thetest should be replaced by the appropriate one-sided limits.


Removable discontinuity

If a function f is said to have a removable discontinuity at a point c inits domain provided that both f (c) and limx→c f (x) = L <∞ exist whilef (x) 6= L.

Removable discontinuity is so named because one can “remove” this pointof discontinuity by redefining f (c) as

f (c) = L = limx→c

f (x).

Each function has a limit as x → 0,and we can remove the discontinuityby setting f (0) equal to this limit.


Jump discontinuity

The function f has a jump discontinuity at c if the one-sided limits existbut have different values.

In the following example, limx→0 f (x) does not exist, and there is no wayto improve the situation by changing f at 0.


Infinite discontinuity

The function f (x) = 1/x2 has an infinite discontinuity.


Oscillating discontinuity

The function f (x) = sin(2π/x) has an oscillating discontinuity. Itoscillates too much to a have a limit as x → 0.


Continuous Functions

A function is continuous on an interval if and only if it is continuous atevery point of the interval.

For example, the semicircle function is continuous on the interval [−2, 2].



A continuous function is one that is continuous at every point of itsdomain.

A continuous function need not be continuous on every interval. Forexample, y = 1/x is not continuous on [−1, 1], but it is continuous overits domain (−∞, 0) ∪ (0,∞).


Properties of Continuous Functions

Theorem 80.

If the functions f and g are continuous at x = c, then the followingcombinations are continuous at x = c.

1. Sums : f + g

2. Differences : f − g

3. Products : f .g

4. Constant multiples : k .f , for any number k

5. Quotients : f /g provided g(c) 6= 0

6. Powers : f r/s , provided it is defined on an open interval containingc, where r and s are integers.


Proof

The above theorem are easily proved from the limit rules.

For instance, to prove the sum property we have

limx→c

{f + g

}(x) = lim

x→c

{f (x) + g(x)

}= lim

x→cf (x) + lim

x→cg(x)

= f (c) + g(c)

= (f + g)(c).

This shows that f + g is continuous.


Proof (using the epsilon-delta definition)


We want to find a positive number δ such that for all x ,

|x − c | < δ implies |f (x) + g(x)− (f (c) + g(c))| < ε.

We have

|f (x) + g(x)− (f (c) + g(c))| ≤ |f (x)− f (c)|+ |g(x)− g(c)|.

From the previous inequality, to get

|f (x) + g(x)− (f (c) + g(c))| < ε

it is observed that we can apply the definition of limit for ε/2 to the limitsf (c), g(c) for the functions f (x), g(x) respectively.


Solution (contd...)

Since limx→c f (x) = f (c), there exists a positive number δ1 such that forall x ,

|x − c | < δ1 implies |f (x)− f (c)| < ε/2.

Similarly, since limx→c g(x) = g(c), there exists a positive number δ2 suchthat for all x ,

|x − c | < δ2 implies |g(x)− g(c)| < ε/2.


Solution (contd...)

Let δ = min{δ1, δ2

}, or any positive number smaller than the minimum.

If |x − c | < δ, then

|x − c| < δ1, so |f (x)− f (c)| < ε/2,

and

|x − c | < δ2, so |g(x)− g(c)| < ε/2.

Therefore

|f (x) + g(x)− (f (c) + g(c))| ≤ ε/2 + ε/2 = ε.

This shows that

limx→c

{f (x) + g(x)

}= f (c) + g(c) = (f + g)(c).


Polynomial and Rational Functions Are Continuous

1. Every polynomial

P(x) = anxn + an−1xn−1 + · · ·+ a0

because

limx→c

P(x) = P(c).

2. If P(x) and Q(x) are polynomials, then the rational functionP(x)/Q(x) is continuous wherever it is defined (Q(c) 6= 0) by thequotient rule.



We proved that

limθ→0

sin θ = 0 and limθ→0

cos θ = 1

using Sandwich Theorem.

The functions y = sin x and y = cos x are continuous at x = 0. In fact,both functions are continuous everywhere.

From the properties of continuous functions, all six trigonometricfunctions (sin x , cos x , tan x , sec x , csc x , cot x) are then continuouswherever they are defined.

For example, y = tan x is continuous on

· · · ∪ (−π/2, π/2) ∪ (π/2, 3π/2) ∪ · · · .


Composites

Composites of continuous functions are continuous.

The idea is that if f (x) is continuous at x = c and g(x) is continuous atx = f (c), then g ◦ f is continuous at x = c .


Composites

Theorem 81 (Composite of Continuous Functions).

If f is continuous at c and g is continuous at f (c), then the compositeg ◦ f is continuous at c.

Intuitively, the above theorem is reasonable because if x is close to c , thenf (x) is close to f (c), and since g is continuous at f (c), it follows thatg(f (x)) is close to g(f (c)).


Composites

The continuity of composities holds for any finite number offunctions. The only requirement is that each function be continuouswhere it is applied.

Example 82.

1. The function y =√

x2 − 2x − 5 is continuous on [0,∞). Note thatthe function y =

√x2 − 2x − 5 is the composite of the polynomial

f (x) = x2 − 2x − 5 with the square root function g(t) =√

t.

2. The function y = x2/3

1+x4 is everywhere continuous.

3. The function y =∣∣∣ x−2x2−2

∣∣∣ is continuous for all x 6= ±√

2.

4. The function y =∣∣∣ x sin xx2+2

∣∣∣ is everywhere continuous.


Continuous Extension to a Point

The function y = sin xx is continuous at every point except x = 0. But we

recall that sin xx has a finite limit 1 as x → 0.

It is therefore possible to extend the function’s domain to include the pointx = 0 in such a way that the extended function is continuous at x = 0.

We define

F (x) =

{sin xx if x 6= 0

1 if x = 0



The function F (x) is continuous at x = 0 because

limx→0

sin x

x= F (0).



More generally, a function (such as a rational function) may have a limiteven at a point where it is not defined. If f (c) is not defined, butlimx→c f (x) = L exists, we can define a new function F (x) by the rule

F (x) =

{f (x) if x is in the domain of f

L if x = c .

The function F is continuous at x = c . It is called the continuousextension of f to x = c .



Example 83.

Show that

f (x) =x2 + x − 2

x2 − 1

has a continuous extension to x = −1 and find that extension.


Solution

Observe that f (x) is not defined at x = 1. If x 6= 1, we have

f (x) =x2 + x − 2

x2 − 1=

x + 2

x + 1.

Note that F (x) = f (x) for all x 6= 1. Also F (x) is defined at x = 1 to be3/2 because

limx→1

x2 + x − 2

x2 − 1= lim

x→1f (x) =

3

2.

Thus F is the continuous extension of f to x = 1.


Intermediate Value Theorem for Continuous Functions

Functions that are continuous on intervals have properties that make themparticularly useful in mathematics and its applications.

One of these is the Intermediate Value Property.

A function is said to have the Intermediate Value Property if wheneverit takes on two values, it also takes on all the values in between.



Theorem 84.

A function y = f (x) that is continuous on a closed interval [a, b] takes onevery value between f (a) and f (b).

In other words, if y0 is any value between f (a) and f (b), then y0 = f (c)for some c in [a, b].



Geometrically, the Intermediate Value Theorem says that any horizontalline y = y0 crossing the y -axis between the numbers f (a) and f (b) willcross the curve y = f (x) at least once over the interval [a, b].

The proof of the Intermediate Value Theorem depends on thecompleteness property of the real number system and hence the proof isomitted.



The continuity of f on the interval is essential to the Intermediate ValueTheorem for continuous functions. If f is discontinuous at even one pointof the interval, the theorem’s conclusion may fail, as it does for thefollowing function.

The function

f (x) =

{2x − 2 if 1 ≤ x < 2

3 if 2 ≤ x ≤ 4

does not take on all values betweenf (1) = 0 and f (4) = 3 ; it misses allthe values between 2 and 3.


Consequence of Intermediate Value Theorem

A Consequence for Graphing : Connectivity

Intermediate Value Theorem for continuous functions is the reason thegraph of a function continuous on an interval cannot have any breaks overthe interval.

It will be connected, a single, unbroken curve, like the graph of sin x .

It will not have jumps like the graph of the greatest integer function orseparate branches like the graph of 1/x .


Consequence of Intermediate Value Theorem

A Consequence for Root Finding

We call a solution of the equation f (x) = 0 a root of the equation or zeroof the function f .

The Intermediate Value Theorem tells us that if f is continuous, then anyinterval on which f changes sign contains a zero of the function.


Exercise

Exercise 85.

In the following exercises, say whether the function graphed is continuouson [−1, 3]. If not, where does it fail to be continuous and why ?

(a) (b)


Solution

(a) No. Discontinuous at x = 2, because the function is not defined atx = 2.

(b) No. Discontinuous at x = 1, because

1.5 = limx→1−

k(x) 6= limx→1+

k(x) = 0.


Exercise

Exercise 86.

Let f (x) be defined by

f (x) =

x2 − 1 if − 1 ≤ x < 0

2x if 0 < x < 1

1 if x = 1

−2x + 4 if 1 < x < 2

0 if 2 < x < 3.


Exercise

Exercise 87.

(a) Does f (−1) exist ?

(b) Does limx→−1+ f (x) exist ?

(c) Does limx→1+ f (x) = f (−1) ?

(d) Is f continuous at x = −1 ?

(e) Does f (1) exist ?

(f) Does limx→1 f (x) exist ?

(g) Does limx→1 f (x) = f (1) ?

(h) Is f continuous at x = 1 ?


Solution

(a) Yes.

(b) Yes, limx→1+ f (x) = 0.

(c) Yes.

(d) Yes.

(e) Yes. f (1) = 1.

(f) Yes. limx→1 f (x) = 2.

(g) No.

(h) No.


Exercise

Exercise 88.

(contd...)

1. (a) Is f defined at x = 27 ? (Look at the definition of f .)(b) Is f continuous at x = 2 ?

2. At what value of x is f continuous ?

3. What value should be assigned to f (2) to make the extended functioncontinuous at x = 2 ?

4. To what new value should f (1) be changed to remove thediscontinuity ?


Solution

1. (a) No.(b) No.

2. [−1, 0) ∪ (0, 1) ∪ (2, 3).

3. f (2) = 0, since limx→2− f (x) = 0 = limx→2+ f (x).

4. f (1) should be changed to

2 = limx→1

f (x).


Exercise

Exercise 89.

At which points do the functions fail to be continuous ? At which points,if any, are the discontinuities removable ? Not removable ? Give reasonsfor your answers.


Solution

Nonremovable discontinuity at x = 1 because limx→1 f (x) fails to exist.

Removable discontinuity at x = 2 by assigning f (2) = 2.


Exercise

Exercise 90.

At what points are the functions continuous ?

1. y = 1x−2 − 3x

2. y = x+3x2−3x−10

3. y = cos xx

4. y = tan πx2

5. y =√

2x + 3

6. y = (2− x)1/5

7.

f (x) =

x3−8x2−4

if x 6= 2, x 6= −2

3 if x = 2

4 if x = −2.


Solution

1. Discontinuous only at x = 2.

2. Discontinuous only at x = 5 or x = −2.

3. Discontinuous only at x = 0.

4. Discontinuous only at x = 2n − 1, n ∈ Z.

5. Continuous on [−32 ,∞).

6. Continuous everywhere

7. Discontinuous only at x = −2.


Exercise

Exercise 91.

Are the functions continuous at the point being approached ?

1. limt→0 sin(π2 cos(tan t)

)2. limy→1 sec(y sec2 y − tan2 y − 1)

3. limt→0 cos(

π√19−3 sec 2t

)


Solution

1. Continuous at t = 0.

2. Continuous at y = 1.

3. Continuous at t = 0.


Exercise

Exercise 92.

1. Define h(2) in a way that extends h(t) = (t2 + 3t − 10)/(t − 2) to becontinuous at t = 2.

2. Define f (1) in a way that extends f (s) = (s3 − 1)/(s2 − 1) to becontinuous at s = 1.


Solution

1. h(2) = 7.

2. f (1) = 32 .


Exercise

Exercise 93.

For what value of b is

g(x) =

{x if x < −2

bx2 if x ≥ −2

continuous at every x ?


Solution

b = −12


Exercise

Exercise 94.

For what value of a is

f (x) =

{a2x − 2a if x ≥ 2

12 if x < 2



Solution

a = 3 or a = −2


Exercise

Exercise 95.

For what values of a and b is

g(x) =

ax + 2b if x ≤ 0

x2 + 3a− b if 0 < x ≤ 2

3x − 5 if x > 2



Solution

a = −32 and b = −3

2 .


Exercise

Exercise 96.

Graph the function f to see whether it appears to have a continuousextension to the origin. If it does, use Trace and Zoom to find a goodcandidate for the extended function’s value at x = 0. If the function doesnot appear to have a continuous extension, can it be extended to becontinuous at the origin from the right or from the left ? If so, what doyou think the extended function’s value(s) should be ?

1. f (x) = sin x|x |

2. f (x) = (1 + 2x)1/x


Solution

1. The function cannot be extended to be continuous at x = 0.

2. The function can be extended : f (0) ≈ 7.39.


Exercise

Exercise 97.

A continuous function y = f (x) is known to be negative at x = 0 andpositive at x = 1. Why does the equation f (x) = 0 have at least onesolution between x = 0 and x = 1 ? Illustrate with a sketch.


Solution

The equation f (x) = 0 has at least one solution between x = 0 and x = 1,by Intermediate Value Theorem.


Exercise

Exercise 98.

Explain why the equation cos x = x has at least one solution.


Solution

By Intermediate Value Theorem, cos x − x = 0 for some x between −π/2and π/2.


Exercise

Exercise 99 (Roots of a cube).

Show that the equation x3 − 15x + 1 = 0 has three solutions in theinterval [−4, 4].


Solution

By Intermediate Value Theorem, f has 3 solutions in (−4,−1), (−1, 1)and (1, 4).


Exercise

Exercise 100 (A function value).

Show that the function F (x) = (x − a)2(x − b)2 + x takes on the value(a + b)/2 for some value of x.


Solution

Since F (a) = a < a+b2 < b = F (b), by Intermediate Value Theorem, there

exists c ∈ (a, b) such that

F (c) =a + b

2.


Exercise

Exercise 101 (Solving an equation).

If f (x) = x3− 8x + 10, show that there are values c for which f (c) equals

(a) π ;

(b) −√

3 ;

(c) 5, 000, 000.


Solution

(a) Since f (1) = 3 < π < 10 = f (0), by Intermediate Value Theorem,there exists c ∈ (0, 1) such that f (c) = π.

(b) Since f (−4) = −22 < −√

3 < 10 = f (0), by Intermediate ValueTheorem, there exists c ∈ (−4, 0) such that f (c) = −

√3.

(c) Since f (0) = 10 < 5, 000, 000 < 999, 999, 010 = f (1000), byIntermediate Value Theorem, there exists c ∈ (0, 1000) such thatf (c) = 5, 000, 000.


Exercise

Exercise 102.

Explain why the following five statemennts ask for the same information.

(a) Find the roots of f (x) = x3 − 3x − 1.

(b) Find the x-coordinates of the points where the curve y = x3 crossesthe line y = 3x + 1.

(c) Find all the values of x for which x3 − 3x = 1.

(d) Find the x-coordinates of the points where the cubic curvey = x3 − 3x crosses the line y = 1.

(e) Solve the equation x3 − 3x − 1 = 0.


Solution

All five statements ask for the same information because of theIntermediate Value Theorem of continuous functions.


Exercise

Exercise 103 (Removable discontinuity).

Give an example of a function f (x) that is continuous for all values of xexcept x = 2, where it has a removable discontinuity. Explain how youknow that f is discontinuous at x = 2, and how know the discontinuity isremovable.


Solution

sin(x−2)x−2 is discontinuous at x = 2, because it is not defined at x = 2.

However, the discontinuity can be removed because f has the limit 1 asx → 2.


Exercise

Exercise 104 (Nonremovable discontinuity).

Give an example of a function g(x) that is continuous for all values of xexcept x = −1, where it has a removable discontinuity. Explain how youknow that g is discontinuous there and why the discontinuity is notremovable.


Solution

g(x) = 1x+1 has a discontinuity at x = −1 because limxto−1 g(x) does not

exist. Note that

limxto−1−

g(x) = −∞ and limxto−1+

g(x) = +∞.


Exercise

Exercise 105 (A function discontinuous at every point).

(a) Use the fact that every nonempty interval of real numbers containsboth rational and irrational numbers to show that the function

f (x) =

{1 if x is rational

0 if x is irrational

is discontinuous at every point.

(b) Is f right-continuous or left-continuous at any point ?


Solution

Suppose x0 is rational. Then f (x0) = 1. Choose ε = 12 . For any δ > 0,

there exists an irrational number x in

(x0 − δ, x0 + δ) =⇒ f (x) = 0.

Then 0 < |x − x0| < δ but |f (x)− f (x0)| = 1 > 12 = ε.

So limx→x0 f (x) does not exist. Hence f is discontinuous at every rationalnumber.

Suppose x0 is irrational. Then f (x0) = 0. Choose ε = 12 . For any δ > 0,

there exists a rational number x in

(x0 − δ, x0 + δ) =⇒ f (x) = 1.

Then 0 < |x − x0| < δ but |f (x)− f (x0)| = 1 > 12 = ε.


Solution (contd...)

So limx→x0 f (x) does not exist. Hence f is discontinuous at everyirrational number.

f is neither right-continuous nor left-continuous at any point x0 because inevery interval (x0 − δ, x0) or (x0, x0 + δ), there are both rational andirrational real numbers.

Thus limx→x−0f (x) and limx→x+

0f (x) do not exist.


Exercise

Exercise 106.

If functions f (x) and g(x) are continuous for 0 ≤ x ≤ 1, could f (x)/g(x)possibly by discontinuous at a point [0, 1] ? Give reasons for your answer.


Solution

Both f (x) = x and g(x) = x − 2 are continuous on [0, 5]. However,f (x)/g(x) is not defined at x = 2, hence it is not continuous at x = 2.


Exercise

Exercise 107.

If the product function h(x) = f (x).g(x) is continuous at x = 0, couldmust f (x) and g(x) be continuous at x = 0 ? Give reasons for youranswer.


Solution

No.

If f (x) = 0 and g(x) = bxc, then h(x) = 0 is continuous at 0, but g(x) isnot continuous at 0.


Exercise

Exercise 108 (Discontinuous composite of continuousfunctions).

Give an example of functions f and g, both continuous at x = 0, for whichthe composite f ◦ g is discontinuous at x = 0. Does this contradict thefollowing theorem ? Give reasons for your answer.

Theorem 109 (Composite of Continuous Functions).

If f is continuous at c and g is continuous at f (c), then the compositeg ◦ f is continuous at c.


Solution

Let f (x) = 1x−1 and g(x) = x + 1. Both functions are continuous at x = 0.

The composition f ◦ g = f (g(x)) = 1x is discontinuous at x = 0.

The theorem on composite of continuous functions requires that f (x) becontinuous at g(0), which is not the case here, since g(0) = 1, and f isundefined at 1.


Exercise

Exercise 110 (Never-zero continuous functions).

Is it true that a continuous function that is never zero on an interval neverchanges sign on that interval ? Give reasons for your answer.


Solution

Yes.

If there is a sign change on an interval, by Intermediate Value Theorem, fhas zero on its interval.


References

1. M.D. Weir, J. Hass and F.R. Giordano, Thomas’ Calculus, 11thEdition, Pearson Publishers.

2. N. Piskunov, Differential and Integral Calculus, Vol I & II (Translatedby George Yankovsky).

3. S.C. Malik and Savitha Arora, Mathematical Analysis, New AgePublishers.

4. R. G. Bartle, D. R. Sherbert, Introduction to Real Analysis, WileyPublishers.


Introduction - National Institute of Technology Karnataka · 2.R. Courant and F.John, Introduction to Calculus and Analysis, Volume II, Springer-Verlag 3.N. Piskunov, Di erential

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