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One-Sided Limits and Continuity Section 2.5 We have already discussed the right hand limit , and the (x) lim xa + f left-hand limit . Let us see the formal definition of (x) lim xa f continuity of a function at a point . (x) f x = a Definition: A function is continuous at a number if the (x) f x = a following conditions are satisfied. 1. is defined. (a) f 2. exists. (x) lim xa f 3. (x) (a) lim xa f = f Geometrically this means the graph of the function does not have a break over/at the point with . x = a Q. At what points is the above function continuous. Soln. . = ,− , } x /{ 8 26 If is not continuous at , then we say that is (x) f x = a (x) f discontinuous at . x = a
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One-Sided Limits and Continuity Section 2biswajit/class6math115.pdf · One-Sided Limits and Continuity Section 2.5 ... Continuity is necessary for the IVT to hold for sure. ... [1,3]

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Page 1: One-Sided Limits and Continuity Section 2biswajit/class6math115.pdf · One-Sided Limits and Continuity Section 2.5 ... Continuity is necessary for the IVT to hold for sure. ... [1,3]

One-Sided Limits and Continuity

Section 2.5

We have already discussed the right hand limit , and the(x)limx→a+

f

left-hand limit . Let us see the formal definition of(x)limx→a−

f

continuity of a function at a point .(x) f x = a

Definition: A function is continuous at a number if the(x)f x = a

following conditions are satisfied.

1. is defined.(a)f

2. exists.(x)limx→a

f

3. (x) (a)limx→a

f = f

Geometrically this means the graph of the function does not have

a break over/at the point with .x = a

Q. At what points is the above function continuous.

Soln. .= − ,− , }x / { 8 2 6

If is not continuous at , then we say that is(x)f x = a (x)f

discontinuous at .x = a

Page 2: One-Sided Limits and Continuity Section 2biswajit/class6math115.pdf · One-Sided Limits and Continuity Section 2.5 ... Continuity is necessary for the IVT to hold for sure. ... [1,3]

is said to be continuous in an interval if is(x)f a, ][ b (x)f

continuous at every point in the given interval.

Q. Is the function (x)  x  if  x =f = { + 2 / 2

              {3  if  x  = 2 Continuous at .x = 2

Soln. f(2) i defined.

exists(x)   limx → 2

f = 2 + 2 = 4

But .(x) = (2)limx → 2

f / f

So the function is discontinuous at x=2. But it is continuous at

all other values of x.

Q. Find all points at which the function is continuous.(x)f = x−2x −42

Soln. The domain of this function is . The function is notx = }{ / 2

continuous at , since it is not defined there. But it isx = 2

continuous at all points in the domain of the function.

Note: as functions. For example is not in(x) =f = x−2x −42 / x + 2 x = 2

the domain of , but it is in the domain of . But at all(x)f x + 2

values of different from 2, the two functions have the samex

value. The graphs of are respectively.,x−2x −42 x + 2

Page 3: One-Sided Limits and Continuity Section 2biswajit/class6math115.pdf · One-Sided Limits and Continuity Section 2.5 ... Continuity is necessary for the IVT to hold for sure. ... [1,3]

Remark: Polynomial functions, rational functions and power

functions are continuous at every point in their respective

domains.

Q. Find the points of discontinuity of the following functions.

+1, , 1, , 3x xx −2x+12x +2x+12  √2x + 1     1x2   2 + 4

Soln. 1}, {x }, }, 0}, }{   <   − 21 { { {

Remark. Except for piecewise defined function, every other

function we will see in this course would be continuous at every

point in their respective domains.

Page 4: One-Sided Limits and Continuity Section 2biswajit/class6math115.pdf · One-Sided Limits and Continuity Section 2.5 ... Continuity is necessary for the IVT to hold for sure. ... [1,3]

In other words if is a number between and . Then theN (a)f (b)f

equation has a solution between and , when f(x) is a(x)f = N a b

continuous function.

Q: Show that the equation has a solution in .x3 + x − 5 = 0 0, ][ 2

Soln. Consider the function . Then , and(x)f = x3 + x − 5 (0) −f = 5

. Since is a continuous function, and ,(2)f = 5 (x)f − 5 < 0 < 5

there must be a between and , such that .c 0 2 (c)f = 0

Please note, IVT may fail, if the function is discontinuous.

Continuity is necessary for the IVT to hold for sure.

Page 5: One-Sided Limits and Continuity Section 2biswajit/class6math115.pdf · One-Sided Limits and Continuity Section 2.5 ... Continuity is necessary for the IVT to hold for sure. ... [1,3]

for this function the equation

has no solution.(x)f = 3

The Intermediate value theorem has the following corollary.

(Existence of zeroes of a continuous function) If is af

continuous function on a closed interval , and if anda, ][ b (a)f

have opposite signs, then there is at least one solution of(b)f

the equation in the interval .(x)f = 0 a, )( b

Geometrically this means, that if the graph of a continuous

function, starts below x-axis at , and goes above x-axis atx = a

, or vice-versa, then it must cross the x-axis at some x = b x = c

Page 6: One-Sided Limits and Continuity Section 2biswajit/class6math115.pdf · One-Sided Limits and Continuity Section 2.5 ... Continuity is necessary for the IVT to hold for sure. ... [1,3]

. This is the root of the equation . But this may failx = c (x)f = 0

if the function is discontinuous.   

But if is not continuous. Then IVT fails.(x)f

Page 7: One-Sided Limits and Continuity Section 2biswajit/class6math115.pdf · One-Sided Limits and Continuity Section 2.5 ... Continuity is necessary for the IVT to hold for sure. ... [1,3]

Q. Show that the function . has a zero in the(x)f = x3 + x − 1

interval . Find an approximate value of this zero.0, )( 1

Soln. is a polynomial function, hence it is continuous(x)f

function. f(0)= -1 < 0 and f(1) = 1 > 0. So by the intermediate

value theorem, f(x) must have at the least one zero in (0,1) .

f(0) = -1, f(1) = 1 , f(x) has a zero in (0,1)

f(0.5) = -0.375, f(x) has a zero in (0.5,1)

f(0.75) = 0.1718 f(x) has a zero in (0.5,0.75)

f(0.625) = -0.13 f(x) has a zero in (0.625,0.75)

f(0.6875)=0.012 f(x) has a zero in (0.625,0.6875)

f(0.65625) = - 0.0611 f(x) has a zero in (0.65625,

0.6875)

f(0.671875) = - 0.02482 f(x) has a zero in (0.671875,0.6875)

f(0.6796875) = - 0.0063138 f(x) has a zero in (0.6796875,0.6875)

Page 8: One-Sided Limits and Continuity Section 2biswajit/class6math115.pdf · One-Sided Limits and Continuity Section 2.5 ... Continuity is necessary for the IVT to hold for sure. ... [1,3]

Hence 0.68359375 0.68 is a good approximation for a zero.≈

This method is called the Method of Bisection. To check f(0.68)

= - 0.005568 . Hence this is a zero upto 2 decimal places.

Q. At what values of , will the following function bek

continuous.

(x)  if  x =f = { x+3x −92 /   − 3

k     if  x{ =   − 3

Soln. This function is continuous at every value of x, except possibly at x=-3. At x=-3, . So the limit6lim

x→ − 3 x+3x −92 = lim

x→ − 3x − 3 =   −  

exists and the function is defined, so the first 2 conditions for

continuity at x = - 3 are satisfied. For the third condition to

also hold good at x = - 3, we must have .k =   − 6

Consider the following graph.

Q. What is the slope of the “secant line” S joining the points (a,f(a)) and (a+h,f(a+h))

Page 9: One-Sided Limits and Continuity Section 2biswajit/class6math115.pdf · One-Sided Limits and Continuity Section 2.5 ... Continuity is necessary for the IVT to hold for sure. ... [1,3]

Soln. .(a+h) − af(a+h) − f(a) =   h

f(a+h) − f(a)

Q. What is the slope of the tangent line” T at the point (a,f(a)) .

Soln. limh → 0 h

f(a+h) − f(a)

Another answer : limx → a x − a

f(x) − f(a)

Q. What is the slope of the tangent line to the graph of at(x)f

the point .x, (x))( f

Soln. limh → 0 h

f(x+h) − f(x)

Q. Find the slope of the tangent line at the point (x,f(x)) for the function .(x) x xf = 2 2 + 5

Soln. We have to find limh → 0 h

f(x+h)−f(x) = limh → 0 h

[2(x+h) +5(x+h)]−[2x +5x]2 2

We use the 4 step process :

Step 1 . (x ) (x ) (x ) (x xh) x h x h xh x hf + h = 2 + h 2 + 5 + h = 2 2 + h2 + 2 + 5 + 5 = 2 2 + 2 2 + 4 + 5 + 5

Step 2. (x ) (x) 2x h xh x h] 2x x]f + h − f = [ 2 + 2 2 + 4 + 5 + 5 − [ 2 + 5

x h xh x h x x h xh h= 2 2 + 2 2 + 4 + 5 + 5 − 2 2 − 5 = 2 2 + 4 + 5

(2h x )= h + 4 + 5

Page 10: One-Sided Limits and Continuity Section 2biswajit/class6math115.pdf · One-Sided Limits and Continuity Section 2.5 ... Continuity is necessary for the IVT to hold for sure. ... [1,3]

Step 3. h xhf(x+h)−f(x) = h

h(2h+4x+5) = 2 + 4 + 5

Step 4. Slope of tangent at (x,f(x)) is ( 2h x )  4x  lim

h → 0  hf(x+h) − f(x) = lim

h → 0+ 4 + 5 =   + 5

Caution: In the above limit is like a constant while is thex h

variable going to .0

Finally : the slope of the tangent line to the graph of at the(x)f

point is .x, (x))( f (x) xf ′ = 4 + 5

Q. Find the slope of the tangent line at the point (1,7) to the graph of .(x) x xf = 2 2 + 5

Soln. Here x=1. But we know slope at (x,f(x)) is 4x+5. So the

slope of the tangent at (1,7) is (1) (1) .f ′ = 4 + 5 = 9

Definition. The derivative of a function is a function ,(x)f (x)f ′

whose value is equal to the slope of the tangent to the graph of

at the point .(x)f x, (x))( f

So .(x)f ′ = limh → 0  h

f(x+h) − f(x)

Q. Find if .(x)f ′ (x) x xf = 2 2 + 5

Page 11: One-Sided Limits and Continuity Section 2biswajit/class6math115.pdf · One-Sided Limits and Continuity Section 2.5 ... Continuity is necessary for the IVT to hold for sure. ... [1,3]

a. Find the slope of the tangent line to the graph of at(x)f

the point 3, 3).( 3

b. Find the equation of the tangent line at the point (3,33).

Soln. .(x) xf ′ = 4 + 5

a. Slope at (3,33) is .(3) 7f ′ = 1

b. The tangent line has slope 17 and passes through (3,33) so

using the point-slope form , it’s equation is (y-33) =

17(x-3) . the standard form is y=17x-18.

Q. If . Find .(x)  f = 1x−2 (x)f ′

Soln. We have to find for limh → 0 h

f(x+h) − f(x) (x) .f = 1x−2

Step 1. (x )f + h = 1(x+h)−2 =

1x+h−2

Step 2. (x ) (x)f + h − f = 1x+h−2 −

1x−2 = (x+h−2)(x−2)

(x−2)−(x+h−2) = x−2−x−h+2(x+h−2)(x−2) =

−h(x+h−2)(x−2)

Step 3. hf(x+h)−f(x) = h

−h(x+h−2)(x−2) = −h

(x+h−2)(x−2)h =−1

(x+h−2)(x−2)

Step 4. . limh → 0 h

f(x+h)−f(x) =   limh → 0

−1(x+h−2)(x−2) =

−1(x−2)2

(We get the above limit by plug-in-g in h=0)

So So (x) −f ′ = 1(x−2)2

Q. Find the equation of the tangent line to the graph of at the point (1,-1).(x) x )f = ( − 2 −1

Soln. The slope of the tangent at this point is . So the(1)f ′ =   − 1

eqn of the tangent is .− ) (x  1)y − ( 1 =   − 1 −  

⇒ y + 1 =   − x + 1  ⇒ y =   − x

Page 12: One-Sided Limits and Continuity Section 2biswajit/class6math115.pdf · One-Sided Limits and Continuity Section 2.5 ... Continuity is necessary for the IVT to hold for sure. ... [1,3]

Definition. The average rate of change of a function on an(x)f

interval is defined to be a, ][ b b−af(b)−f(a)

The ( instantaneous ) rate of change of a function at , is(x)f x = a

defined to be equal to .(a)f ′

Q. For the function . Calculate the average rate of(t)  f = 1t−2

change of the function over the interval [1,3]. Also find the

instantaneous rate of change at t=1.

Soln. Av rate of change over the interval [1,3] =

/2 .3−1f(3)−f(1) = 2

1−(−1) = 2 = 1

Instantaneous rate of change at t=1 = (1) . f ′ = −1(1−2)2 =   − 1

Q. Let the position of a car at the time be given by thet

function (t) 2t t) meters.s = ( 2 + 5

a. Find the velocity and acceleration of the car as a function

of time .t

b. Find the velocity and acceleration at sec.t = 1

Soln. We have found earlier (t) t . s′ = 4 + 5

a. The velocity of the car is (instantaneous)rate of change of

position of the car, so .(t) (t) tv = s′ = 4 + 5

The acceleration of the car is the (instantaneous) rate of

change of velocity of the car, so . Note we have not(t) (t)a = v′ = 4

found , it is an exercise for the students.(t)v′

b. In particular velocity of the car at issect = 1

(1) (1) m/secv = 4 + 5 = 9

And the acceleration of the car at sec is .t = 1 (1) m/seca = 4 2

Page 13: One-Sided Limits and Continuity Section 2biswajit/class6math115.pdf · One-Sided Limits and Continuity Section 2.5 ... Continuity is necessary for the IVT to hold for sure. ... [1,3]