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Continuity & Differentiability Return to Main Page

Part B: Differentiability
Exercises for This Topic
miscellaneous on-line topics Index of On-Line Topics
Everything for Calculus
for Everything for Finite Math
Everything for Finite Math &
Calculus Applied to the Real Calculus
World Utility: Function Evaluator &
Grapher
Part A: Continuity
Part A: Continuity

Note To understand this topic, you will need to be familiar with limits, as discussed in
the chapter on derivatives in Calculus Applied to the Real World. If you like, you can
review the topic summary material on limits or, for a more detailed study, the on-line
tutorial on limits.

To begin, we recall the definition (see Section 6 in the derivatives chapter in Calculus
Applied to the Real World.) of what it means for a function to be continuous at a point
or on a subset of its domain.

Continuous at a Point
The function f is continuous at the point a in its domain if:
1. lim f(x) exists;
x a
2. lim f(x) = f(a)
x a
If f is not continuous at a, we say that f is discontinuous at a.

Note
If the point a is not in the domain of f, we do not talk about whether or not f is
continuous at a.

Continuous on a Subset of the Domain

The function f is continuous on the subset S of its domain if it continuous at
each point of S.

Examples
1. All closed form functions are continuous on their (whole) domain. A closed-
form function is any function that can be obtained by combining constants,
powers of x, exponential functions, radicals, logarithms, and trigonometric
functions (and some other functions we shall not encounter in this text) into a
single mathematical formula by means of the usual arithmetic operations and
composition of functions. Examples of closed form functions are:

2x
3x2 - x + 1, , and ex2-1.
x+3
 2

2. The function f(x) = 1/x is continuous at every point of its domain. (Note that 0
is not a point of the domain of f, so we don't discuss what it might mean to be
continous or discontinuous there.)

3. The function

-1 if x 2
f(x) =
x +x if x > 2
2

is not a closed-form function (since we need two algebraic formulas to specify

it). Moreover, it is not continuous at x = 2, since limx 2f(x) does not exist.

Example 1 Recognizing Points of Discontinuity in a Graph

Let f have the graph shown below.

At which points of the domain of f is the function f discontinuous?

Solution

According to the definition, f can fail to be continuous at a point a of its domain if

either:

1. lim f(x) does not exist, or

x a
2. lim f(x) exists, but f(a)
x a

Looking at the figure, we see that the possible points where things might go wrong are
at x = -1, 0, 1, and 2. Let us look at these points one-at-a-time.
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x = -1: Notice that x = -1 is a point of the domain of f, and lim x

-1f(x) does not exist, because the left- and right-habd limits
disagree. Therefore, f has a discontinuity at x = -1.

x = 0: From the graph, we see that f(0) is not defined. Therefore,

0 is not in the domain of f, so we cannot say that f has a
discontinuity at x = 0. (Some authors would say that f is
discontinuous at x = 0, but we don't consider such points...)

x = 1: At the point where x = 1, the limit does exist, however,

lim f(x) = 1, but

x 1
f(1) = 2
Thus, since the limit at 1 does not equal the value of the function, we have a
discontinuity at x = 1.

x = 2: Although there is a cusp in the graph at x = 2, you will see that the limit at x = 2
exists, and agrees with f(2). Therefore, f is continuous at x = 2.

Thus, the only points of discontinuity in the domain of f occur at x = -1 and x =1.

Here is a more interactive example.

Example 2 Recognizing Points of Discontinuity in a Graph

Let f have the graph shown below.

Q f is at x = -2
Q f is at x = -1
Q f is at x = 1
Q f is at x = 2

The next examples discuss functions specified algebraically rather than geometrically.
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Example 3 Identifying Points of Discontinuity in a Non-Closed Form

Function

Let f be specified as

x2+4 if x 0
f(x) = x + 1 if 0 < x 1 .
x2 + 1 if x > 1
Then, the only conceivable points of discontinuity occur at x = 0 and x = 1. (At all other
points in the domain, f can be specified by a single closed-form formula; for example,
near x = 0.5, f can be specified as x + 1.)

Looking near x = 0, we get

lim f(x) = 4 Substitute x = 0 in the first (closed-form) formula

x 0-
lim f(x) = 1 Substitute x = 0 in the second (closed-form) formula
x 0+

Since these limits disagree, limx 0f(x) does not exist, and so the function is
discontinuous at x = 0.

If you now compute the left- and right limits at x = 1, on the other hand, you will fined
they agree and equal 2, which is also the value of f(1). Thus, f is continuous at x = 1,
and so the only point of discontinuity is x = 0.

Example 4 Fixing up a Function

Let f be given by

x2 - x + 1 if x 1
f(x) = .
kx2 + 1 if x > 1
For which value(s) of k is f continuous on its domain?

Solution

Since the only place where something can go wrong is at x = 1, we look at the left- and
right limits, as well as the value of the function.

lim f(x) = 1
x 1-
lim f(x) = k+1
x 1+
f(1) = 1
To make the left- and right limits agree, we must make k = 1 (and this also makes f(1)
agree with those limits). Thus, we take k = 1.
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You can now either go on and try those exercises that deal with continuity in the
exercise set for this topic, or first finish the on-line text by going to Part B:
Differentiability.

Return to Main Page

Part B: Differntiability
Exercises for This Topic
Index of On-Line Topics
Everything for Calculus
Everything for Finite Math
Everything for Finite Math & Calculus
Utility: Function Evaluator & Grapher
Last Updated:October, 1999
Copyright © 1999 StefanWaner and Steven R. Costenoble

CONTINUITY OF FUNCTIONS OF ONE VARIABLE

The following problems involve the CONTINUITY OF A FUNCTION OF ONE

VARIABLE. Function y = f(x) is continuous at point x=a if the following three
conditions are satisfied :

i.) f(a) is defined ,

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ii.) exists (i.e., is finite) ,

and

iii.) .

Function f is said to be continuous on an interval I if f is continuous at each point x in I.

Here is a list of some well-known facts related to continuity :

1. The SUM of continuous functions is continuous.

2. The DIFFERENCE of continuous functions is continuous.

3. The PRODUCT of continuous functions is continuous.

4. The QUOTIENT of continuous functions is continuous at all points x where

the DENOMINATOR IS NOT ZERO.

5. The FUNCTIONAL COMPOSITION of continuous functions is continuous

at all points x where the composition is properly defined.

6. Any polynomial is continuous for all values of x.

7. Function ex and trigonometry functions and are continuous for all

values of x.

Most problems that follow are average. A few are somewhat challenging. All limits are
determined WITHOUT the use of L'Hopital's Rule. If you are going to try these
problems before looking at the solutions, you can avoid common mistakes by using the
above step-by-step definition of continuity at a point and the well-known facts, and by

giving careful consideration to the indeterminate form during the computation of

limits. Knowledge of one-sided limits will be required. For a review of limits and
indeterminate forms click here.

o PROBLEM 1 : Determine if the following function is continuous at x=1 .

SOLUTION 1 : Function f is defined at x=1 since

i.) f(1) = 2 .
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The limit

= 3 (1) - 5

= -2 ,

i.e.,

ii.) .

But

iii.) ,

so condition iii.) is not satisfied and function f is NOT continuous at x=1 .

o PROBLEM 2 : Determine if the following function is continuous at x=-

2.

SOLUTION 2 : Function f is defined at x=-2 since

i.) f(-2) = (-2)2 + 2(-2) = 4-4 = 0 .

The left-hand limit

= (-2)2 + 2(-2)

=4-4

=0.

The right-hand limit

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= (-2)3 - 6(-2)

= -8 + 12

=4.

Since the left- and right-hand limits are not equal, ,

ii.) does not exist,

and condition ii.) is not satisfied. Thus, function f is NOT continuous at x=-2 .

Click HERE to return to the list of problems.

o PROBLEM 3 : Determine if the following function is continuous at x=0 .

SOLUTION 3 : Function f is defined at x=0 since

i.) f(0) = 2 .

The left-hand limit

=2.

The right-hand limit

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=2.

Thus, exists with

ii.) .

Since

iii.) ,

all three conditions are satisfied, and f is continuous at x=0 .

PROBLEM 4 : Determine if the function is continuous at

x=-1 .

SOLUTION 4 : Function h is not defined at x=-1 since it leads to division by zero. Thus,

i.) h(-1)

does not exist, condition i.) is violated, and function h is NOT continuous at x = -1 .

o PROBLEM 5 : Check the following function for continuity at x=3 and

x=-3 .

SOLUTION 5 : First, check for continuity at x=3 . Function f is defined at x=3 since

i.) .

The limit
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(Circumvent this indeterminate form by factoring the numerator and the denominator.)

(Recall that A2 - B2 = (A-B)(A+B) and A3 - B3 = (A-B)(A2+AB+B2 ) . )

(Divide out a factor of (x-3) . )

i.e.,

ii.) .

Since,

iii.) ,

all three conditions are satisfied, and f is continuous at x=3 . Now, check for continuity
at x=-3 . Function f is not defined at x = -3 because of division by zero. Thus,

i.) f(-3)
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does not exist, condition i.) is violated, and f is NOT continuous at x=-3 .

o PROBLEM 6 : For what values of x is the function

continuous ?

SOLUTION 6 : Functions y = x2 + 3x + 5 and y = x2 + 3x - 4 are continuous for all

values of x since both are polynomials. Thus, the quotient of these two functions,

, is continuous for all values of x where the denominator, y = x2 +

3x - 4 = (x-1)(x+4) , does NOT equal zero. Since (x-1)(x+4) = 0 for x=1 and x=-4 ,
function f is continuous for all values of x EXCEPT x=1 and x=-4 .

o PROBLEM 7 : For what values of x is the function

continuous ?

SOLUTION 7 : First describe function g using functional composition. Let f(x) = x1/3 ,

, and k(x) = x20 + 5 . Function k is continuous for all values of x since it is a

polynomial, and functions f and h are well-known to be continuous for all values of x .
Thus, the functional compositions

and

are continuous for all values of x . Since

function g is continuous for all values of x .

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o PROBLEM 8 : For what values of x is the function

continuous ?

SOLUTION 8 : First describe function f using functional composition. Let g(x) = x2 - 2x

and . Function g is continuous for all values of x since it is a polynomial,

and function h is well-known to be continuous for . Since g(x) = x2 - 2x = x(x-2) ,

it follows easily that for and . Thus, the functional composition

is continuous for and . Since

function f is continuous for and .

o
o
o
o
o
o
o
o

o PROBLEM 9 : For what values of x is the function

continuous ?

SOLUTION 9 : First describe function f using functional composition. Let

and . Since g is the quotient of polynomials y = x-1 and y =

x+2 , function g is continuous for all values of x EXCEPT where x+2 = 0 , i.e.,
 13

EXCEPT for x = -2 . Function h is well-known to be continuous for x > 0 . Since

, it follows easily that g(x) > 0 for x < -2 and x > 1 . Thus, the functional
composition

is continuous for x < -2 and x > 1 . Since

function f is continuous for x < -2 and x > 1 .

o PROBLEM 10 : For what values of x is the function

continuous ?

SOLUTION 10 : First describe function f using functional composition. Let

and h(x) = e x , both of which are well-known to be continuous for all

values of x . Thus, the numerator is continuous (the functional

composition of continuous functions) for all values of x . Now consider the denominator

. Let g(x) = 4 , h(x) = x2 - 9 , and . Functions g and h are

continuous for all values of x since both are polynomials, and it is well-known that
function k is continuous for . Since h(x) = x2 - 9 = (x-3)(x+3) = 0 when x=3 or x=-

3 , it follows easily that for and , so that

is continuous (the functional composition of continuous

functions) for and . Thus, the denominator is

continuous (the difference of continuous functions) for and . There is
one other important consideration. We must insure that the DENOMINATOR IS
NEVER ZERO. If

then
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Squaring both sides, we get

16 = x2 - 9

so that

x2 = 25

when

x = 5 or x = -5 .

Thus, the denominator is zero if x = 5 or x = -5 . Summarizing, the quotient of these

continuous functions, , is continuous for and , but

NOT for x = 5 and x = -5 .

o PROBLEM 11 : For what values of x is the following function

continuous ?

SOLUTION 11 : Consider separately the three component functions which determine f .

Function is continuous for x > 1 since it is the quotient of continuous

functions and the denominator is never zero. Function y = 5 -3x is continuous for

since it is a polynomial. Function is continuous for x < -2 since

it is the quotient of continuous functions and the denominator is never zero. Now check
for continuity of f where the three components are joined together, i.e., check for
continuity at x=1 and x=-2 . For x = 1 function f is defined since

i.) f(1) = 5 - 3(1) = 2 .

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The right-hand limit

(Circumvent this indeterminate form one of two ways. Either factor the numerator as
the difference of squares, or multiply by the conjugate of the denominator over itself.)

=2.

The left-hand limit

= 5 - 3(1)

=2.

Thus,

ii.) .

Since

iii.) ,

all three conditions are satisfied, and function f is continuous at x=1 . Now check for
continuity at x=-2 . Function f is defined at x=-2 since

i.) f(-2) = 5 - 3(-2) = 11 .

The right-hand limit

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= 5 - 3( -2)

= 11 .

The left-hand limit

= -1 .

Since the left- and right-hand limits are different,

ii.) does NOT exist,

condition ii.) is violated, and function f is NOT continuous at x=-2 . Summarizing,

function f is continuous for all values of x EXCEPT x=-2 .

o PROBLEM 12 : Determine all values of the constant A so that the

following function is continuous for all values of x .

SOLUTION 12 : First, consider separately the two components which determine

function f . Function y = A2 x - A is continuous for for any value of A since it is a
polynomial. Function y = 4 is continuous for x < 3 since it is a polynomial. Now
determine A so that function f is continuous at x=3 . Function f must be defined at x=3 ,
so

i.) f(3)= A2 (3) - A = 3 A2 - A .

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The right-hand limit

= A2 (3) - A

= 3 A2 - A .

The left-hand limit

=4.

For the limit to exist, the right- and left-hand limits must exist and be equal. Thus,

ii.) ,

so that

3 A2 - A - 4 = 0 .

Factoring, we get

(3A - 4)(A + 1) = 0

for

or A = -1 .

For either choice of A ,

iii.) ,

all three conditions are satisfied, and f is continuous at x=3 . Therefore, function f is

continuous for all values of x if or A = -1 .

o PROBLEM 13 : Determine all values of the constants A and B so that the

following function is continuous for all values of x .
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SOLUTION 13 : First, consider separately the three components which determine

function f . Function y = Ax - B is continuous for for any values of A and B
since it is a polynomial. Function y = 2x2 + 3Ax + B is continuous for for
any values of A and B since it is a polynomial. Function y = 4 is continuous for x > 1
since it is a polynomial. Now determine A and B so that function f is continuous at x=-1
and x=1 . First, consider continuity at x=-1 . Function f must be defined at x=-1 , so

i.) f(-1)= A(-1) - B = - A - B .

The left-hand limit

= A (-1) - B

=-A-B.

The right-hand limit

= 2(-1)2 + 3A(-1) + B

= 2 - 3A + B .

For the limit to exist, the right- and left-hand limits must exist and be equal. Thus,

ii.) ,

so that

2A - 2B = 2 ,

or

(Equation 1)

A-B=1.

Now consider continuity at x=1 . Function f must be defined at x=1 , so

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i.) f(1)= 2(1)2 + 3A(1) + B = 2 + 3A + B .

The left-hand limit

= 2(1)2 + 3A(1) + B

= 2 + 3A + B .

The right-hand limit

=4.

For the limit to exist, the right- and left-hand limits must exist and be equal. Thus,

ii.) ,

or

(Equation 2)

3A + B = 2 .

Now solve Equations 1 and 2 simultaneously. Thus,

A - B = 1 and 3A + B = 2

are equivalent to

A = B + 1 and 3A + B = 2 .

Use the first equation to substitute into the second, getting

3 (B + 1 ) + B = 2 ,

3B+3+B=2,

and

4 B = -1 .

Thus,
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and

For this choice of A and B it can easily be shown that

iii.)

and

iii.) ,

so that all three conditions are satisfied at both x=1 and x=-1 , and function f is
continuous at both x=1 and x=-1 . Therefore, function f is continuous for all values of x

if and .

o PROBLEM 14 : Show that the following function is continuous for all

values of x .

SOLUTION 14 : First describe f using functional composition. Let g(x) = -1/x2 and h(x)
= ex . Function h is well-known to be continuous for all values of x . Function g is the
quotient of functions continuous for all values of x , and is therefore continuous for all
values of x except x=0 , that x which makes the denominator zero. Thus, for all values
of x except x=0 ,

f(x) = h ( g(x) ) = e g(x) = e -1/x2

is a continuous function (the functional composition of continuous functions). Now

check for continuity of f at x=0 . Function f is defined at x=0 since

i.) f(0) = 0 .
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The limit

(The numerator approaches -1 and the denominator is a positive number approaching

zero.)

so that

=0,

i.e.,

ii.) .

Since

iii.) ,

all three conditions are satisfied, and f is continuous at x=0 . Thus, f is continuous for all
values of x .

o PROBLEM 15 : Let
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Show that f is continuous for all values of x . Show that f is differentiable

for all values of x, but that the derivative, f' , is NOT CONTINUOUS at
x=0 .

SOLUTION 15 : First show that f is continuous for all values of x . Describe f using

functional composition. Let , , and k(x) = x2 . Function h is

well-known to be continuous for all values of x . Function k is a polynomial and is
therefore continuous for all values of x . Function g is the quotient of functions
continuous for all values of x , and is therefore continuous for all values of x except
x=0 , that x which makes the denominator zero. Thus, for all values of x except x=0 ,

is a continuous function (the product and functional composition of continuous

functions). Now check for continuity of f at x=0 . Function f is defined at x=0 since

i.) f(0) = 0 .

The limit does not exist since the values of oscillate between -1

and +1 as x approaches zero. However, for

so that

Since

it follows from the Squeeze Principle that

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ii.) .

Since

iii.) ,

all three conditions are satisfied, and f is continuous at x=0 . Thus, f is continuous for all

values of x . Now show that f is differentiable for all values of x . For we can

differentiate f using the product rule and the chain rule. That is, for the derivative
of f is

Use the limit definition of the derivative to differentiate f at x=0 . Then

.
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Use the Squeeze Principle to evaluate this limit. For

If , then

If , then

In either case,

and it follows from the Squeeze Principle that

Thus, f is differentiable for all values of x . Check to see if f' is continuous at x=0 . The
function f' is defined at x=0 since

i.) f'(0) = 0 .

However,

ii.)

does not exist since the values of oscillate between -1 and +1 as x approaches
zero. Thus, condition ii.) is violated, and the derivative , f' , is not continuous at x=0 .

NOTE : The continuity of function f for all values of x also follows from the fact that f
is differentiable for all values of x .
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Click HERE to return to the original list of various types of calculus

problems.

Your comments and suggestions are welcome. Please e-mail any correspondence to
Duane Kouba by clicking on the following address :

kouba@math.ucdavis.edu

• About this document ...

Duane Kouba
1998-06-01