Math1401 Intro to Limits and Continuity

7/31/2019 Math1401 Intro to Limits and Continuity

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SETON HALL UNIVERSITY

Introduction toLIMITS AND CONTINUITY

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PROF. MICHAEL ARYEE 2 SETON HALL UNIVERSITY

Consider the graphs of the two function below.

.1,1

1)(and1)(

32

x

x

xxgxxxf


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The limits technique provides us with the

tools we need in order to predictthe value or

height of the function by using the values of

the independent variable nearsome given

point, while completely ignoring the value of

the function at that point.

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So, how can we estimate (or guess) the value of thefunction

when x = 1?

First, we can investigate the behavior of the graph ofg

nearx= 1 by using the two sets of values of x:

(1) one set that approaches 1 from theleftof x = 1

x-Values From thelefttowards 1: {0, 0.5, 0.9, 0.99, 0.999}

(2) one set that approaches 1 from therightof x = 1

x-Values From therighttowards 1:{2, 1.5, 1.1, 1.01, 1.001}

1,1

1)(

3

x

x

xxg


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The answer can be found by numerically investigating the

behavior of the graph ofgas x gets closer and closer to 1 fromthe left and from the right.

Does g(x) appear to get closer and closer to a fixed number?

Question: What

happens to g(x) asx gets closer to 1?

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From the table, we see that when x is close to 1(that is, on either side of 1), g(x) is close to 3.

In fact, it appears that we can make the values ofg(x) asclose as we like to 3 by taking xsufficiently close to 1.For example we can consider x = 0.9999 and 1.0001

We can therefore express the above investigation by

saying that the limit of the function g(x) as x

approaches 1 is equal to 3.


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The above example shows how to estimate thevalue of a function numerically using the concept oflimits.

The notation for the above example is:

We will introduce the informal definition of limits andthen show more examples of the above.

limx

x

x

1

31

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Consider a function fwith an independent variable

xdefined on an open interval (a, b), except for somefixed numberc, which also belongs to the interval,

then the function fis said to approach L(the limit),as xapproaches c, if the value of f(x) get closer and

closer to a unique number L as values of x getscloser and closer to c.

Using symbols we write:

lim ( )x c

f x L


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Note that cand L are both numbers and they satisfy

the following conditions:

Wheneverxis close to c, but not equal to c, f(x) isclose to L.

As x gets closer and closer to c, but not equal to c,f(x) gets closer and closer to L.

We can make f(x) close to L as we please bychoosing xto be close to cbut not equal to c.

We want the difference between f(x) and L to be asnear zero as possible for all values ofxdefined onthe open interval (a, b) containing c.

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When we consider the limit from only one direction,

either from the left only, or from the right only, the

answer is regarded as a one-sided limit.

A one-sided limit can be a Left-hand limit or aRight-hand limit.

We denote a Left-Hand Limit with a little negativesymbol (-) at the location where the exponentof thenumber should be. We write:

We denote a Right-Hand Limit with a little positivesymbol (+) at the location where the exponentof thenumber should be. We write:

lim ( )x c

f x

lim ( )x c

f x


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Using our table below:

Left Handg x

x

:lim ( )

1

3Right Hand

g xx

:

lim ( )1

3

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In general, the limit of f(x) existsifallof the followingstatements are true:

1. is definite and has a value.

2. is definite and has a value.

3 =

lim ( )x c

f x

lim ( )x c

f x

lim ( )

x cf x

lim ( )x c

f x

if and only if andlim ( )x c

f x L

lim ( )

x cf x L

lim ( )x c

f x L

In General:


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1.

2.

3.

lim ( )x

g x

1

3

lim ( )x g x

1 lim ( )x g x

1 3

lim ( )x

g x

1

3

if and only if andlim ( )x

g x

1

3lim ( )x

g x

1

3 lim ( )x

g x

1

3

That is:

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Lets consider the function whose graph is

represented below.

If we inspect the graph, we can see that there is a

breakat x= 4.


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If we travel towards x= 4 from the right, we will

arrive at the height of 2.

.

The height at which we arrive at if we are traveling

from the RIGHT is called the Right-Hand Limit of f(x)as xapproaches a specific value ofx, in this case asxapproaches 4.

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If we travel towards x= 4 from the left, we will arriveat the height of 1.

The height at which we arrive at if we are traveling

from the LEFT is called the Left-Hand Limit off(x)as xapproaches a specific value ofx, in this case asxapproaches 1.


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D.N.E. means Does Not Exist.

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To estimate the limit, there are three approaches. We

have just illustrated two of these approaches

(numerical approach and graphical approach). The

three approaches are listed below:

Numerical approach (limit obtained by weconstructing a table of values)

Graphical approach (limit obtained by sketching agraph)

Analytic approach (limit obtained by usingalgebra)


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Find by means of a table of values.

limx

xx

1

2 1

1

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limx

x

x

1

21

1

From the table:


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Alternatively, you can setup a horizontal table like the

one below and determine the limits from both sides.

limx

x

x

1

2 1

1

x 0 0.5 0.9 0.99 0.999 1 1.001 1.01 1.1 1.5 2

1 1.5 1.9 1.99 1.999 2.001 2.01 2.1 2.5 3

limx

x

x

1

21

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Given the graph off(x) shown below, Determine thelimit at x = 0.


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Given the graph off(x) shown below, Determine thelimit at x = 1.

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This involves using algebra to find the limit

To find the limit of a function using the analytical

approach, first try direct substitution.

We usually use direct substitution when the function

you are dealing with is a polynomial, a rational, or

any algebraic function whose domain do not exclude

the specific value that x approaches.

Direct substitution is always valid for polynomials and

rational functions with nonzero denominators.

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Iffis a polynomial function, then

Ex.

Ifris a rational function given by

and then

Ex.

lim ( ) ( )x a

p x p a

1-(1)-2(1)3(1)1x2x3xlim2424

1

x

0)( ag

r x f xg x

( ) ( )( )

lim ( ) ( )x a

r x r a

lim( ) ( )

( )x

x x

x

2

2 22

2

2 2 2

2 2

4

41


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If the limit of f(x) as x approaches c cannot beevaluated through direct substitution, because thiswill cause denominator to be zero, then try tofactorand cancel out the common factors, and thenusedirect substitution again.

, Note that Direct Substitution =

Now factor, cancel, and do direct substitution again:

0

0lim

x

x x

x

3

2 6

3


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If the numerator approaches 0 and the denominatoralso approaches 0 and the numerator involves asquare root expression, thenrationalize thenumerator by multiplying both the numerator and

the denominator by the conjugate of the

numerator.

In general if f(x) = g(x) for every x except possibly x =c, in some interval containing c, then

lim ( ) lim ( ) ( )x a x a

f x g x g a

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Find

Multiply numerator and denominator by

conjugate .

limx

x

x

4

2

4

2x

.

4

1

22

1

24

1

2

1lim

4

xx


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Find

.

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Given the function

determine whether or not the limit existat x=1.

.

lim ( ) lim ( )x x

f x x

1 1

3 3 1 3

lim ( ) . .x

f x D N E1

Left:

Right: lim ( ) lim ( )x xf x x 1 1 2 3 2 1 3 5

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Determine whether or not the function f(x) = | x 2 |

has a limit at x = 2.

The function f(x)= | x 2 | is the same as:

.

lim ( ) lim[ ( )] ( )x x

f x x

2 2

2 2 2 0

lim ( )x

f x2

0

Left:

Right: lim ( ) lim( ) ( )x x

f x x

2 2

2 2 2 0


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SETON HALL UNIVERSITY

CONTINUITY


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A function fis continuous at a point x = cif thegraph of fhas no gap, break, split, holes, jumps,or a missing point for f(x) at the point x = c.

In addition, a function fis continuous at a point x = cif you can move along the graph offthrough thepoint (c, f(c)) with a pencil without lifting the pencilfrom the paper.

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A function f is continuous at a point x = c if all of the

following THREE conditions are fulfilled:

1. f(c ) is defined

2. exists, and

3. = f(c ).

For a function to be continuous at the point x = c, the functionmust have a value at x = c(which isf(c)) and the limit offas xapproaches c( ) must have a value which is the sameasthe value of the function at x = c

lim ( )x c

f x

lim ( )x c

f x

lim ( )x c

f x


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A function is said to be discontinuousat x = c, if ahole, gap, or break occurs in the graph at x = c,meaning the function violates one of the three

items above. Below are some examples of

discontinuities at x = c.

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When a function is not defined at certain values,these values are certainly the points of discontinuityof the function.

Because division by zero is not allowed in

mathematics, any number or numbers that when

substituted into the function causes thedenominator to be zero, thus making the function

undefined, represent points of discontinuity.

There are two main categories of discontinuity:

removable and non-removable.

Non-removable has two divisions: a jumpdiscontinuity and an infinite discontinuity.


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A jump discontinuity

A non-removable jump discontinuity is thesituation where we are unableto find a value for

because

lim ( )x c

f x

lim ( ) lim ( )x c x c

f x f x

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An infinite discontinuity

A non-removable infinite discontinuity is thesituation where we are unableto find a value for

because = infinity.

lim ( )x c

f x

lim ( )x c

f x


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A removable discontinuity is the situation where

#1) We are able to find a value for , but we are

unableto find a value forf(c),

or

#2) we are able to get values for

both and f(c), however,

their values do not match,

that is, lim ( ) ( )x c

f x f c

lim ( )x c

f x

lim ( )x c

f x

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Given the function

Determine if the function is continuous at x=1.

1. Find f(c):

2. Find 2a.

2b.

3. Compare #1. and #2.

Therefore, the function is continuous at x =1.

.

lim ( ) lim ( )x x

f x x

1 1

3 3 1 3

lim ( )x

f x1

3

Left:

Right: lim ( ) lim ( )x x

f x x

1 1

2 1 2 1 1 3

f( ) ( )1 3 1 3

lim ( )x f x1

lim ( ) ( )x

f x f

1

1 3


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Given the function

determine if the function is continuous at x=1.

1. Find f(c):

2. Find

2a.

2b.


Therefore, the function is DISCONTINUOUS at x=1.

.

lim ( ) lim ( )x x

f x x

1 1

3 1 3 1 1 4

lim ( )x

f x DNE1

Left:

Right: lim ( ) lim ( )x x

f x x

1 1

2 3 2 1 3 5

f( ) ( )1 3 1 1 4 lim ( )x

f x1

lim ( ) ( )x

f x f

1

1

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Given the function

Determine if the function is continuous at x=-3.

1. Find f(c):

2. Find


Therefore, the function is CONTINUOUS at x=-3.

.

f( ) 3 5

lim ( ) ( )x

f x f

3

3 5

lim ( )x f x 3


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Given the function

Determine the value of a that makes f(x) continuous

We need to find the value of a so that :

(continue on next slide)

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Math1401 Intro to Limits and Continuity

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