limits and continuity - Thomas Jeffersalittle/cs_math/lim.pdf · 2005-05-13 · LIMITS AND CONTINUITY update 3/12/2005 _____ Part 1: Finding Limits Numerically and Graphically We

1

LIMITS AND CONTINUITYupdate 3/12/2005

______________________________________________________________________________ Part 1: Finding Limits Numerically and Graphically We begin the discussion of limits by revisiting rational functions and examining their graphs and tables of values. Example 1: Consider first the graph of ( ) 3f x x= + . The table of values of the function is shown below as well as the graph.

Notice that as the x values approach 3 for the function ( )f x , the y values approach 6 and eventually reach 6.

We write ( )3

lim 3 6x

x→

+ = .

Consider a second function 2 9

( )3

xg x

x−=−

If you look at the graph of g(x), you would see no difference.

An examination of the table reveals a hole at 3x = . A better graph of the function would indicate a hole at x = 3.

Just as before, we can say 2

3

9lim 6

3x

xx→

− =−

So, the limit of a function need not be a defined value of the function.

Intuitively, when we say lim ( )x c

f x L→

= , we mean that as x gets close to c, then f(x) gets close to a limit of L.

We can find limits numerically (by examining tables of values), graphically (by drawing and tracing along the graph of a function), and analytically (by utilizing the rules of algebra and theorems defining the rules of limits). In this first section, we will concentrate on finding limits numerically and graphically.

2

Example 2: Find 0

limx

x

x→

Examine both the table and graph for this function:

Again, the graphing calculator does not indicate the holes that occur at 0x = . A better graph is

As x approaches 0 from the left, the limit appears to be -1. We say0

lim 1x

x

x−→= − .

As x approaches 0 from the right, the limit appears to be 1. We say0

lim 1x

x

x+→= .

Note: We use a “+” sign to indicate approaching from the right and we use a “-” to indicate approaching from the left.

In order for a limit to exist, the limit must be a single value that the function approaches. Therefore, we say 0

limx

x

x→

does not exist since 0 0

lim limx x

x x

x x− +→ →≠

Example 3: Find 3

2lim

3x

xx→

+−

Examine both the table and graph for this function:

As x approaches 3 from the left, the limit is growing in the negative direction without bound.

We say 3

2lim

3x

xx−→

+ = −∞−

3

As x approaches 3 from the right, the limit is growing in the positive direction without bound.

We say 3

2lim

3x

xx+→

+ = +∞−

Because there is not a single value (or direction) that the function approaches, we say 3

2lim

3x

xx→

+−

does not exist.

We can expand our understanding of limits to say,

If lim ( ) lim ( )x c x c

f x f x L− +→ →

= = , then lim ( )x c

f x L→

= . In other words, if the "left-hand limit" equals the "right-and

limit, then the limit of the function exists and is the common limit.

Example 4: Find 20

1limx x→

Examine both the table and graph for this function:

As x approaches 0 from the left, the limit is growing in the positive direction without bound. As x approaches 0 from the right, the limit is again growing in the positive direction without bound. Even though there is not a single

numerical value that the function approaches, there is a single direction. Therefore, we say 20

1limx x→

= ∞ .

Note: The symbol “�” means increases without bound. The symbol “- �” means decreases without bound.

Example 5: Find 0

coslim

1000x

xx

→

� �−� �� �

When you graph this function, the limit appears to be 0.

4

When you examine the table of values, it is obvious that not only is the limit of the function -0.001, but also that the value of the function exists at this point and is this value.

Therefore, 0

cos 1lim

1000 1000x

xx

→

� �− = −� �� �

Example 6: Find 0

1lim sinx x→

� �� �� � � �� �

Examine the table and graph of this function.

A cursory examination might lead to the conclusion that the limit of this function is 0. But upon closer examination,

it becomes obvious that the graph is oscillating so wildly that there is not a single limit. Therefore, 0

1lim sinx x→

� �� �� � � �� �

does not exist. Notice that the limit of a function does not exist if (1) the same numerical value is not approached from both the left and right hand side of the given value; (2) the same infinite direction is not approached from both the left and right hand side of the given value; and, (3) the function oscillates so rapidly that a single value is not approached. ______________________________________________________________________________ Homework Exercises Part 1: Finding Limits Numerically and Graphically Find each of the following limits numerically and/or graphically.

1. 2

5

25lim

5x

xx→

−−

4. 2

0limx

xx→

2. 3

1

1lim

1x

xx→

−−

5. 4 3 2

21

3 13 27 36lim

3 4x

x x x xx x→

+ − − ++ −

3. 2

43

12limx

xx→

− 6.

0

3 3limx

xx→

+ −

5

Find each of the following limits numerically and/or graphically.

7. 0

1 coslimx

xx→

− 11. 20

sinlimx

xx→

8. ( )23

1lim

3x x→ − 12.

1

2 1lim

2 1x

xx→

+−

9. 2

3lim

2x x→ − 13.

0

5 25limx

xx→

− −

10. 0

sinlimx

xx→

14. 0

1lim cosx x→

� �� �� � � �� �

15. In the exercises below, the graph of a function is given. State whether or not the

3lim ( )x

f x→

exists and, if it

does, give its value.

0

1

1

a.

0

1

1

b.

0

1

1

c.

0

1

1

d.

16. Find the following one-sided limits (if they exist):

a. 25

5lim

25x

xx+→

−−

b. 22

2lim

4x

xx+→

−−

c. 4

lim2x

xx+→ −

d. 4

2lim

4x

xx−→

−−

6

17. For the function f graphed below, find the indicated limit or function value, or state that the limit does not

exist. a.

3lim ( )x

f x→−

b. ( 3)f − c. 1

lim ( )x

f x→−

d. ( 1)f − e. (1)f f. 2

lim ( )x

f x−→

g. 2

lim ( )x

f x+→

h. 2

lim ( )x

f x→

i. (2)f

1

1

18. Sketch the graph of the piecewise function below and then find each of the following limits or function

values. If the limit does not exist, write "DNE"; if the value is not defined write "undefined".

2

1 1

( ) 1 1 2

5 2

x if x

g x x if x

x if x

− + <��= − < <��

− ≥��

a.

1lim ( )x

g x→

b. (1)g

c.

2lim ( )x

g x→

d. (2)g

19. Sketch the graph of a function f that satisfies all of the following conditions: a. Its domain is the interval [0, 4] b. (0) (1) (2) (3) (4) 1f f f f f= = = = =

c. 1

lim ( ) 2x

f x→

=

d. 2

lim ( ) 1x

f x→

=

e. 3

lim ( ) 2x

f x→

=

20. As we have seen, we can use tables to approximate the value of a limit. The smaller the change in x (∆x) we choose, the better our approximation should be. We also say that the left- and right-hand limits must be

equal, in order for the limit to exist. Use your calculator to find the following limits.

a. 1

1lim

1x

xx→

−−

b. 0

1 coslimx

xx→

−

c. 0

sin(2 tan )limx

xx→

d. 2

2

4lim

2x

xx+→

−−

e. 100

lim logx

x x+→� f.

0lim x

xx

+→

7

21. (OPTIONAL) The following program (written for the TI-83) computes 12 values for x to the left and to the right of the given c value. The successive ∆x-s are 0.1, 0.01, . . . 0.0000000000001. For example, if

you were to use the LIMIT program to approximate 0

sin 2lim

3x

xx→

, enter sin 2

3x

x in Y1 and run the

program. The screen should appear as follows:

Note that the program computes both the right and left hand limit, and if this value is the same, the program

concludes that the limit of the function is this single value. In this case the limit is 23

. If these values are

not sufficient for determining the limit, then the computation is beyond the capacity of the calculator. The calculator will give the message “after 12 approximations, no limit”. In addition, for a function such as lncot

xx

the restrictions on the domain will not allow the calculations. You will receive an error message.

Methods that we will study in Part 2 will be necessary to ascertain these limits. �

������� ���� ��������� ����������������������� ���������������������������� ������� �� ���!�"����#$ ��%&��'����(�$ ���������!�"����#$ ��%&��'����)�$ �����* �� ���� �����"��� +�'����,�

…(Program Continued ) '���� ����,�������-.���� +�-���/!���������/�����'����0����(��������'1����������������)��������.�������������"����#���%� ��* ��'����2�#� �%3+��4����������5�������4�'����0����2����������/�5��/���!5������0�

Use the LIMIT program to approximate the following limits, if possible.

a. 1

1lim

1x

xx→

−−

b. 0

1 coslimx

xx→

−

c. 0

sin(2 tan )limx

xx→

d. 2

2

4lim

2x

xx+→

−−

e. 100

lim logx

x x+→� f.

0lim x

xx

+→

8

______________________________________________________________________________ Part 2: Finding Limits Analytically Studying the graph of a function can sometimes be misleading in determining the limit of a function. It is possible to find limits analytically using limit theorems and algebraic techniques. The most basic theorem governing analysis of limits is the Principal Limit Theorem. Principal Limit Theorem: Let c and k be real numbers, n > 0, and let f and g be functions with limits at c such that

lim ( )x c

f x L→

= and lim ( )x c

g x M→

= , then

1. Constant Rule: limx c

k k→

=

2. Identity Rule: limx c

x c→

=

3. Coefficient Rule: lim ( ) lim ( )x c x c

kf x k f x kL→ →

= =

4. Sum Rule: [ ]lim ( ) ( ) lim ( ) lim ( )x c x c x c

f x g x f x g x L M→ → →

+ = + = +

5. Difference Rule: [ ]lim ( ) ( ) lim ( ) lim ( )x c x c x c

f x g x f x g x L M→ → →

− = − = −

6. Product Rule: [ ]lim ( ) ( ) lim ( ) lim ( )x c x c x c

f x g x f x g x L M→ → →

= =� � �

7. Quotient Rule: lim ( )( )

lim( ) lim ( )

x c

x cx c

f xf x Lg x g x M

→

→→

� �= =

� � provided lim ( ) 0

x cg x

→≠

8. Power Rule: [ ]lim ( ) lim ( )nn n

x c x cf x f x L

→ →� �= =� �

It is important to be able to state each of these rules in English. For instance, Rule 1 could be stated, “The limit of a constant function n is the constant.” Rule 4 could be stated, “The limit of a sum is the sum of the limits.” etc. Consider how these rules are used to evaluate the following limit.

Example 1: Find 31

2 1lim

2x

xx→

−−

Solution: 31

2 1lim

2x

xx→

−−

= ( )( )

13

1

lim 2 1

lim 2x

x

x

x→

→

−

− Quotient Rule

= 1 13

1 1

lim 2 lim1

lim lim 2x x

x x

x

x→ →

→ →

−

− Difference Rule

=

( )1 1

3

1 1

2lim lim1

lim lim 2

x x

x x

x

x

→ →

→ →

−

−

= ( )3

2(1) 1

1 2

−−

Constant and Identity Rules

= 1−

Coefficient Rule Power Rule

9

Example 2: Find 2

0lim cosx

x x→

� . This example uses our rules as well as our knowledge of trigonometric graphs.

Solution: 2 2

0 0 0lim cos lim limcosx x x

x x x x→ → →

=� � Product Rule

( )2

0 0lim lim cosx x

x x→ →

= � Power Rule

= 20 1� Identity Rule and Trigonometric Graph = 0 Note that both of these limits could have been found by simply substituting the value of c into the limit statement. This is not true for all types of functions, but is true for two significant groups of functions.

Substitution Theorem: If f is a polynomial function or rational function then lim ( ) ( )x c

f x f c→

= provided that if f is

a rational function the value of the denominator does not equal 0.

Example 3 (Substitution): Find 5 3 2

42

8 6 3 4 1lim

3 2 7x

x x x xx x→

− + − +− +

.

Solution: Because this is a rational function whose denominator does not equal 0 when x = 2 is substituted, we may apply the Substitution Theorem:

5 3 2 5 3 2

4 42

8 6 3 4 1 8(2) 6(2 ) 3(2 ) 4(2) 1 213lim

3 2 7 3(2 ) 2(4) 7 47x

x x x xx x→

− + − + − + − += =− + − +

The exact limit would have been difficult to obtain graphically. It is sometimes necessary to simplify expressions algebraically before applying the Substitution Theorem.

Example 4 (Cancellation): Find 2

3

6lim

3x

x xx→

− −−

.

Solution: The Substitution does not apply because the denominator equals 0 when 3 is substituted, But, if the expression is simplified by using cancellation, the theorem can be applied.

2

3 3 3

6 ( 3)( 2)lim lim lim( 2) 5

3 3x x x

x x x xx

x x→ → →

− − − += = + =− −

Example 5 (Rationalization): Find 4

4lim

2x

xx→

−−

.

Solution: It is first necessary to rationalize the expression.

( ) ( ) ( )4 4 4

4 24 2lim lim lim 2 4

42 2x x x

x xx xx

xx x→ → →

− +− + = = + =−− +

�

�

10

______________________________________________________________________________ Homework Exercises Part 2: Finding Limits Analytically 1. Write the 8 rules of the Principal Limit Theorem in English. Use the Principal Limit Theorem to find each of the following limits. Justify each step with a named rule.

2. ( )12

lim 5 7x

x→

+ 3. 4

3lim

3x

xx→ −

Find each of the following limits analytically.

4. 3

3

27lim

3x

xx→

−−

5. 2

25

5lim

25x

x xx→

−−

6. 89 56 18

59 46 151

2 6 8 7lim

3 17 19x

x x x xx x x→−

+ − − −+ − +

7. 4

limx

e→

8. ( )4

41

1lim

1x

x

x→

−−

9. 0

2 4limx

xx→

− +

10. 3 2

3 20

3 2 7lim

8 9 7x

x x xx x x→

− + −+ − +

11. 22

21

lim4x

xx→

−

−

Draw the graph of a function f such that lim ( )

x cf x

→ does not exist when c has the following values

12. 1, 2, 3 13. -2, 0, 2 For Exercises 14 - 19, find the limits if lim ( ) 2

x cf x

→= and lim ( ) 3

x cg x

→= −

14. lim 2 ( ) 4 ( )x c

f x g x→

− 15. [ ]3lim ( ) 1x c

f x→

+

16. 2 ( ) 3 ( )

lim( ) ( )x c

f x g xg x f x→

+−

17. [ ]lim ( ) 2 ( )x c

f x g x→

�

18. ( )lim ( ) ( ) 3x c

f x g x→

+� �� �� 19. 2( )

lim1 ( )x c

f xg x→ −

11

______________________________________________________________________________ Part 3: Limits at Infinity and Asymptotes There are two important definitions involving limits and infinity that are directly related to the asymptotes of rational function graphs. A horizontal asymptote of a rational function can be determined by using the following definition.

Limit at infinity: lim ( )x

f x L→∞

= means that as x increases without bound the function approaches a numeric

value L.

L

L

A vertical asymptote of a rational function can be determined by using the following definition.

Infinite Limit: lim ( )x c

f x→

= ∞ means that as x approaches the value c, the function increases without bound.

c

The dashed lines are called asymptotes. Informally a line is called an asymptote of a curve if the curve comes arbitrarily close to the line. Formally, asymptotes are defined with respect to limits at infinity and infinite limits:

Horizontal asymptotes; limits at infinity: The horizontal line y L= is an asymptote to the graph of f if and only if

lim ( ) lim ( )x x

f x L or f x L→∞ →−∞

= = .

Vertical asymptotes; infinite limits: The vertical line x c= is an asymptote to the graph of f if and only if

lim ( ) lim ( ) lim ( ) lim ( )x c x c x c x c

f x or f x or f x or f x− − + +→ → → →

= +∞ = −∞ = +∞ = −∞ .

12

There are 12 different situations involving limits and infinity as illustrated below: 1. lim ( )

x cf x

−→= ∞ 2. lim ( )

x cf x

−→= −∞ 3. lim ( )

x cf x

+→= ∞

c

c

c

4. lim ( )

x cf x

+→= −∞ 5. lim ( )

x cf x

→= ∞ 6. lim ( )

x cf x

→= −∞

c

c

c

7. lim ( )x

f x L→∞

= 8. lim ( )x

f x L→−∞

= 9. lim ( )x

f x→∞

= ∞

L

L

10. lim ( )

xf x

→−∞= ∞ 11. lim ( )

xf x

→∞= −∞ 12. lim ( )

xf x

→−∞= −∞

In order to find a vertical asymptote, look for the roots of the denominator, i.e., where the denominator is undefined.

Recall that if a 00

situation occurs, there is a hole at that point. If a 0

number situation occurs, there is a vertical

asymptote.

13

In order to find horizontal or vertical asymptotes consider the following four statements and accompanying graph.

1. 1

lim 0x x→∞

= 2. 1

lim 0x x→−∞

=

3. 0

1limx x+→

= ∞ 4. 0

1limx x−→

= −∞

To find the limits at infinity, divide each term in the rational expression by the highest power of x and use the statements 1 - 4 above.

Example 1: Find 3

3 2

3 4 2lim

2 1x

x xx x→∞

− ++ −

.

Solution: The highest power is 3x . Divide each term by this power.

3

3 3 3 3 3 3

3 2 23 2

3 3 3 3 3

3 4 2 4 233 4 2

lim lim lim2 1 12 1 2

x x x

x x xx x x x x x x

x x xx xx x x x x

→∞ →∞ →∞

− + − +− + = =+ − + − + −

Using the Principal Limit Theorem and the fact that 1

lim 0x x→∞

= , the last statement can be

simplified and the limit found to be 32

.

Example 2: Find 4

3 2

3 4 2lim

2 1x

x xx x→∞

− ++ −

.

Solution: In this case when each term is divided by the highest power, the result is

3 3

2

3 3

4 23

lim2 1x

xx xx

x x x

→∞

− +

+ −

Each term in the denominator is approaching 0, but the numerator approaches infinity. Therefore, the limit of the entire expression is ∞ .

14

Example 3: Find the horizontal and vertical asymptotes for the function ( )5 3

xf x

x−=

+

Solution: You learned in Algebra 2 that in order to find the vertical asymptote, set 5 3 0x + = and solve.

35

x = − is the vertical asymptote for this function. What you are actually doing is to find c for

lim ( )x c

f x→

= ∞ . To find the horizontal asymptote, find lim ( )x

f x→∞

. It is 15

− . Thus, there is a

horizontal asymptote at 15

y = − . Confirm your findings by examining the graph of the function.

[-4.7, 4.7] by [-1.6, 1.7]

Example 4: Find the horizontal and vertical asymptotes of 2

1( )

1x

f xx

+=+

Solution: There are no real solutions to 2 1 0x + = . Therefore, there are no vertical asymptotes. lim ( ) 0x

f x→∞

= . Therefore, 0y = is the horizontal asymptote. The graph is pictured below.

[-10, 10] by [-2, 2] In Examples 3 and 4, the degree of the numerator was less than or equal to the degree of the denominator. In both cases, it is possible to find a horizontal asymptote. But, if the degree of the numerator exceeds the degree of the denominator, an oblique asymptote occurs.

15

Example 5: Find the asymptotes of 2 1

( )x

f xx+=

Solution: The function has a vertical asymptote at 0x = . When we find lim ( )

xf x

→∞, we find it to be ∞ .

But we can indicate the direction that the curve takes by finding the oblique asymptote. Since 2 1 1x

xx x+ = + , the end behavior of the graph approximates the line y x= . Thus, y x= acts as an

oblique asymptote for the graph.

[-5, 5] by [-6, 6] ______________________________________________________________________________ Homework Exercises Part 3: Limits at Infinity and Asymptotes Find each of the following limits.

1. 2

2

2lim

2 5x

xx→∞

++

2. 4 3

4 2

2 2 1lim

7 3 4x

x x xx x x→∞

+ − +− + +

3. 3

3

3000lim

1000x

xx x→∞ −

4. 5

4

4 2 3lim

1 2x

x xx x→∞

− +− +

5. 2

4

2 3lim

5 2 1x

x xx x→∞

+− +

6. 21

lim1x

xx→∞

−+

7. Let 2 2

2 2

2 5 12( ) and ( )

2 3 2 3x x k x x k

f x g xx x x x

+ + + += =− − − −

. Can the number k be found so that

a.

3lim ( )x

f x→

exists? Justify your answer. d. 3

lim ( )x

g x→

exists? Justify your answer.

b.

1lim ( )x

f x→−

exists? Justify your answer. e. 1

lim ( )x

g x→−

exists? Justify your answer.

c. lim ( )

xf x

→∞exists? Justify your answer. f. lim ( )

xg x

→∞exists? Justify your answer.

16

Write a limit statement that describes the pictured situation. 8. 9.

1

1

1

1

10. 11.

1

1

1

1

12. Sketch the graph of a function ( )f x that satisfies all of the following conditions:

• ( )f x has domain ( 2, )− ∞

• ( )f x has range ( , 4)−∞

• ( 1) 0; (0) 0; (6) 0f f f− = = =

• 2

lim ( )x

f x+→−

= −∞

• 4

lim ( ) 2x

f x→

= −

• lim ( ) 4x

f x→∞

=

13. Sketch the graph of a function ( )f x that satisfies all of the following conditions:

• ( )f x has domain of all real numbers

• ( )f x has range ( 4,4)−

• 4

lim ( ) 2x

f x→

= −

• lim ( ) 4x

f x→∞

= −

• lim ( ) 4x

f x→−∞

=

14. Give examples of polynomials f(x) and g(x) such that lim ( )x

f x→∞

= ∞ , lim ( )x

g x→∞

= ∞ and

17

a. ( )

lim( )x

f xg x→∞

= ∞ b. ( )

lim 0( )x

f xg x→∞

=

Find all horizontal, vertical and/or oblique asymptotes for the following functions.

15. 2

2 5( )

4x

f xx

+=−

16. 2 2 1

( )2

x xf x

x− +=

+

17. 2

2

5 3( )

9x x

f xx

− +=−

18. 4 2

2

2 3( )

4x x x

f xx

− − +=−

19. Does 2

2

2 5 19

x xy

x− +=

− cross its horizontal asymptote? If so, at what point?

______________________________________________________________________________ Part 4: Continuity A closely-related concept to limits is continuity. Informally when we say a function is continuous, we mean that the graph is unbroken or that it can be drawn without lifting the pencil from the paper. Each of the following functions are continuous over the set of real numbers.

( )f x x= ( ) sing x x= ( ) 2 1h x x= +

The following graphs are not continuous (are discontinuous) over the set of real numbers.

2 9

( )3

xp x

x−=−

( )x

q xx

= 2

( )3

xr x

x+=−

To formally state why these functions are not continuous, consider the following definition:

18

The function f is continuous at c if and only if

1. f is defined at c 2. f has a limit as x approaches c, and 3. the limit equals f(c)

In other words, lim ( ) ( )x c

f x f c→

=

Therefore, to prove each of the first three functions (f, g, and h) is continuous at a given point, all three conditions must be demonstrated.

Example 1: Prove ( )f x x= is continuous at x = 0.

Solution: (1) (0) 0 0f = =

(2) 0

lim 0x

x→

=

(3) 0

(0) lim 0x

f x→

= =

Example 2: Prove 2 9

( )3

xp x

x−=−

is not continuous (is discontinuous) at x = 3.

Solution: (1) (3)p is undefined. There is a hole in the graph at this point. It is not possible

for the limit to equal an undefined value. Therefore, p(x) is discontinuous at x = 3.

The function p(x) is said to have a “removable” discontinuity at x = 3 because the function can be made continuous by appropriately defining the function at the discontinuity; i.e., if we define (3) 6p = , i.e.,

2 93

( ) 3

6 3

xif x

p x x

if x

−� ≠= � −� =�

The functions q(x) and r(x) on the previous page do not have removable discontinuities.

Example 3: Let 2 2 1

( )1

ax if xf x

x if x

+ < −�= �≥ −��

What value of a makes f continuous at x = -1?

Solution: In order for f(x) to be continuous at x = -1,

1 1lim ( ) lim ( ) ( 1)

x xf x f x f

− +→− →−= = − . Since

( 1) 1f − = − and 1

lim 1x

x−→−

= − , we must guarantee that 2

1lim 2 1

xax

+→−+ = −

Since 2 2ax + is a polynomial function, we can substitute -1 for x and solve the resulting equation.

2( 1) 2 1 2 1 3a a a− + = − � + = − � = −

So, 23 2 1

( )1

x if xf x

x if x

− + < −�= �≥ −��

and the graph of f(x) is as pictured below.

19

Several functions that we have studied are continuous. Polynomial functions are continuous.

A rational function is continuous at each member of its domain, that is, except where its denominator is zero.

The absolute value function is continuous.

If n is odd the nth root function is continuous at every real number c; if n is even, it is continuous at every positive real number c.

If f and g are continuous functions at a point c, then so are kf, f + g, f - g, f . g, fg

(as long as

g(c) � 0), and f n.

If g is continuous at c, and f is continuous at g(c), then f(g) is continuous at c; i.e.,

lim ( ( )) ( ( ))x c

f g x f g c→

=

Example 4: Show that 2( ) 2 3 4P x x x= − + is continuous.

Solution: Let ( )f x x= and 2( ) 2 3 4g x x x= − + . f(x) is continuous because it is the absolute value

function. g(x) is continuous because it is a polynomial function. ( ) ( ( ))P x f g x= . Thus, P(x) is a composite function of two continuous functions and is therefore continuous.

Example 5: Study the graph below and determine where the function is discontinuous. Identify the removable discontinuities and define the value of the function in order to remove the discontinuity.

20

1

1

Solution: The graph is discontinuous at -2, 0, and 2. The discontinuity at 2 is removable. If you let (2) 1f = the discontinuity is removed.

______________________________________________________________________________ Homework Exercises Part 4: Continuity Prove each of the following functions is continuous at the given value, using the three part definition of continuity at a point c.

1. ( ) 2 3f x x at c= − = 2. ( ) 4f x x at c= =

3. 1

( ) 1f x x at cx

= − = 4. 2( ) 1f x x at c= = −

Prove each of the following is discontinuous at the given value

5. 1

( ) 1 0f x at cx

= − = 6. 2

2 2( ) 2

2

x if xf x at c

x x

− >�= =�� ≤�

7. 2 4

( ) 22

xf x at c

x−= =−

Explain how you know that the given function is continuous over its domain.

8. ( ) 2f x x= − 9. ( )f x x x= �

State whether or not the following functions are continuous at x = 1

10. 3( ) 2 4 5f x x x= − + 11. 2

( )1

xf x

x=

−

12. 2 2

( )1

xf x

x−=−

13. 3 1

( )1

xf x

x−=−

14.

3 11( ) 1

3 1

xfor xf x x

for x

−�� ≠= � −�

=��

In Exercises 15 - 20, the given function has a removable discontinuity at a certain point. How should the function be defined in order to make it continuous everywhere?

21

15. 2 4

( )2

xf x

x−=+

16. 24 9

( )2 3x

f xx

−=−

17. 4

( )2

xf x

x−=−

18. sin

( )x

f xx

=

19. 1

( )1

xf x

x−=

− 20.

2 1( )

1x

f xx

−=−

21. Given two functions ( )f x and ( )h x such that 3 2( ) 3 4 12f x x x x= − − + and

( )3( ) 3

3

f xfor xh x x

p for x

� ≠= � −�

=�

a. Find all zeros of the function f. b. Find the value of p so that the function h is continuous at x = 3. Justify your answer. c. Using the value of p found in (b), determine whether h is an even function. Justify your answer. (This problem is from the 1976 AP AB Calculus Exam)

22. Given the function f defined by 2

2 2( )

2x

f xx x

−=+ −

a. For what values of x is f(x) discontinuous? b. At each point of discontinuity found in part (a), determine whether f(x) has a limit and, if so, give the value of the limit. c. Write an equation for each vertical and horizontal asymptote. Justify your answer. d. Draw a detailed and complete graph of f(x) using the information you found in parts (a), (b), and (c). (This problem is from the 1978 AP AB Calculus Exam) 23. Suppose that f is continuous at x = 3 and (3) 4f = − . Evaluate

3lim ( ).x

f x→

24. Suppose that f is continuous at x = 2 and that 2

lim ( ) 25x

f x→

= . Indicate which of the statements about f must

be true, which might be true, and which cannot be true. Justify your answers. a. 2 is in the domain of f. b. (2) 25f =

c. 2

lim ( ) 25x

f x−→

=

25. Sketch the graph of a function f that satisfies all of the following conditions:

• Its domain is [-3, 3].

22

• ( 3) ( 1) (1) (3) 2f f f f− = − = = = • It is discontinuous at -1 and 1. • f is continuous on the open interval (-1, 1).

26. Suppose that the function f is continuous at x = 5 and that f is defined by the rule

2 2 5( )

4 7 5

kx if xf x

x if x

+ <�= �+ ≥��

a. Find k. b. Find

5lim ( ).x

f x→

27. Let f be the pictured graph. Using the graph, evaluate each of the following and then identify all points of discontinuity.

1

1

a.

3lim ( )

xf x

−→− b.

3lim ( )

xf x

+→− c.

3lim ( )x

f x→−

d. ( 3)f − e.

1lim ( )

xf x

−→− f.

1lim ( )

xf x

+→−

g.

1lim ( )x

f x→−

h. ( 1)f − i. 1

lim ( )x

f x−→

j.

1lim ( )x

f x+→

k. 1

lim ( )x

f x→

l. (1)f

m.

3lim ( )x

f x+→

n. 3

lim ( )x

f x−→

o. 3

lim ( )x

f x→

p. (3)f q. What are the points of discontinuity? ______________________________________________________________________________ Homework Exercises Part 5: Review

23

True/False. Determine if each statement is true or false. If the statement is false, give an example of why it is false. 1. If lim ( )

x cf x L

→= then ( ) .f c L=

2. If ( )f c is not defined then lim ( )

x cf x

→ does not exist.

3. If ( )f x is a polynomial, then lim ( ) ( ).

x cf x f c

→=

4. If lim ( ) lim ( )

x c x cf x f x

− +→ →= then f is continuous at .x c= .

5. If ( ) ( )f x g x≠ for all x, then lim ( ) lim ( )x c x c

f x g x→ →

≠

6. lim ( ) 1x

f x→∞

= − implies that the graph of f has a vertical asymptote.

Find the indicated limit or state that it does not exist.

7. 3

2

8lim

2x

xx→−

++

8. 32

2lim

8x

xx→−

++

9. 20

coslimx

xx→

10. 3 5

lim4x

xx→∞

−+

11. ( )( )3 2

2 2

2 5 3lim

3 5x

x xx x→∞

− +− +

12. 1

1lim

1x

xx→

−+

13. 1

1lim

1x

xx→

−−

14. 13

1 3lim

1 3x

x

x−→

−−

15. 2

2 5lim

1x

xx→∞

−−

16. 2 1

lim2x

xx→∞

++

17. Sketch the graph of a function f that satisfies all of the following conditions:

• Its domain is [-1, 5]. • ( 1) (1) (3) (5) 2f f f f− = = = = −

• f is continuous except at 1x =

• 1 1

lim ( ) 2 lim ( ) 0x x

f x and f x− +→ →

= − =

24

18. Find each of the following values if

22 1

( ) 2 1 1

2 1

x if x

f x x if x

x if x

< −��= − ≤ ≤��

− >��

a. ( 1)f − b. (1)f

c.

1lim ( )

xf x

−→− d.

1lim ( )

xf x

+→−

e. 1

lim ( )x

f x→−

f. Where is f(x) discontinuous?

19. Which of the following statements are true of the function graphed here?

1

1

a.

1lim ( ) 2x

f x+→

= b. 2

lim ( )x

f x→

does not exist

c.

2lim ( ) 3x

f x→

= d. 2

lim ( ) 3x

f x−→

=

e. 1

lim ( ) 2x

f x−→

= f. 1

lim ( )x

f x→

does not exist

g.

2 2lim ( ) lim ( )

x xf x f x

+ −→− →−= does not exist h. ( )f x is continuous at 2x =

i. ( )f x is continuous at 2x = − j. lim ( )

x cf x

→ exists at every c in (-3, -2)

20. If a continuous, increasing function f is described by the table below, find a. lim ( )

xf x

→∞ b.

3lim ( )x

f x→

x 1 2 2.9 3.1 5 7 9 11 15

f(x) 5 7 7.9 8.1 20 24 24.7 24.9 24.99

25

21. Use your calculator to estimate sin

limx

xx→∞

22. For the function ( )( ) 5 1 xf x e−= − , find

a. lim ( )

xf x

→∞ b.

0lim ( )x

f x→

23. Use the calculator to find the ( )1

0lim 1 xx

x→

+ . This is an approximation of an important value. What is this value

and the values of the variations in the table below?

1

0lim 1

1

x

x

x→

� �+� �� �

2

0lim 1

2

x

x

x→

� �+� �� �

1

lim 1x

x x→∞

� �+� �� �

3

lim 1x

x x→∞

� �+� �� �