3-4 the Limit of a Function

8/13/2019 3-4 the Limit of a Function

http://slidepdf.com/reader/full/3-4-the-limit-of-a-function 1/17



3.3 The Limit of a Function MHR 141

We can also determine these values using the TABLE feature of a graphing calculator orgraphing software, after selecting Ask mode on the TABLE SETUP screen:

From the table and the graph of f (x), note that when x is close to 2, on either side of 2, f (x)is close to 12. Furthermore, it appears that, as x approaches 2 more and more closely, thevalues of f (x) approach 12 more and more closely. Verify this by entering values even closer

to 2, such as 1.999999 and 2.000001, into the graphing calculator. This is expressed as

“the limit of as x approaches 2 is equal to 12,”

or, in brief,

In general, the limit of a function is written , read as “the limit of f (x),

as x approaches a, equals L,” if the values of f (x) approach L more and moreclosely as x approaches a more and more closely, from either side of a, but x a.

Note the phrase “but x a” in the limit definition. This means that, in finding the limit of f (x) as x approaches a, we never consider x a. In fact, f (x) need not even be defined when

x a, as in above. The only thing that matters is how f (x) behaves

near a. The reason for this is that we want limits to be applicable to the calculation of

the slope of a tangent to a curve at a point P, by first determining the slope of asecant PQ and then letting Q approach P. But, as we saw in the previous two sections,the slope of a secant is undefined when Q P, so we need to exclude Q P from ourreasoning.

The graph of the function has a hole at x 2 because f (x) is not

defined there. This hole is called a discontinuity, and the function f (x) is described as beingdiscontinuous at x 2. There are three different types of discontinuities: removable, jump,

and infinite. The discontinuity in f (x) is called removable, because we could remove itby defining f (x) appropriately for the value x 2. We discuss the other two types of discontinuities later in this section.

The figure on the next page shows the graphs of three similar functions that behavedifferently at x a. In part a), f (a) is defined at a and f (a) L. In part b), f (a) is definedat a and f (a) L. In part c), f (a) is not defined at a. But in each case, despite the differencesin the values of f (a), we have .lim ( )

x af x L

→=

f x x x x

x( ) =

+ − −−

3 2 4 4

2

f x x x x

x( ) =

+ − −−

3 2 4 4

2

lim ( )x a

f x L→

=

limx

x x x

x→

+ − −−

=2

3 2 4 4

212

x x x

x

3 2 4 4

2+ − −

−

Window variables:x [0, 4], y [0, 24]



142 MHRChapter 3

Investigate & Inquire: Discontinuities in Functions

In this investigation, you will explore discontinuities in a specific function and examine thebehaviour of the function values near these discontinuities.

1. a) Using a graphing calculator or graphing software, graph .

Use the decimal (“friendly”) window by pressing 4.b) Alternatively, sketch a graph of f (x) for values of x close to 2, on either side of 2.Repeat for values of x close to 4.c) Describe what happens when x 4 and x 2. Is the function defined at these values

of x? How can you tell from the equation of the function?2. What kind of discontinuity occurs at x 4?

3. Construct tables by finding the values of for values of x

approaching 4 from both sides, but not equal to 4. Alternatively, use the TABLE feature of

a graphing calculator or graphing software. Why can x not be equal to 4?

4. a) Find .

b) Suppose f (x) is redefined as

How does this affect the limit calculated in part a)?

5. Factor the function and simplify it. Call the new factored function

g (x). Describe the difference between f (x) and g (x).

6. Construct tables by finding the values of for values of x

approaching 2 from both sides, but not equal to 2. Alternatively, usethe TABLE feature of a graphing calculator or graphing software. Why can xnot be equal to 2?

7. How does the behaviour of the function at x 2 differ from that at x 4?

f x xx x

( ) = −− +6 82

f x x

x x( ) = −

− +4

6 82

f x

x

x x x

x

( )

.

=

−

− + ≠

=

4

6 84

0 5 4

2if

if

limx

x

x x→

−

− +4 2

6 8

f x x

x x( ) = −

− +6 82

f x x

x x( ) = −

− +4

6 82

0 x

y

L

a

a)

0 x

y

L

a

b)

0 x

y

L

a

c)

y f x= ( ) y f x= ( ) y f x= ( )



3.3 The Limit of a Function MHR143

In the investigation, the function f has an infinite discontinuity at x 2. We have already

seen such discontinuities—for example, the function has an infinite discontinuity

at x 0, where the graph also has a vertical asymptote. Discontinuities involving verticalasymptotes will be discussed in detail in Chapter 6.

Another type of discontinuity is often encountered in electric circuits that are turned onand off, in economics problems, and in other situations. Consider the net worth ofa company or business that is experiencing financial difficulty. The value of the company isdecreasing over time until the net worth is less than zero. Eventually the company declaresbankruptcy. At this time, the company’s debts are forgiven, its net worth is zero, and so ithas a chance to make a new start. If the new business is successful, the net worth of thecompany could rise rapidly. A function that could be used to model this situation is

where t is the time, in years, and v is the net worth of the small business, in tens of thousands of dollars. The discontinuity at t 2 is called a jump discontinuity, which occurswhen there is a break in the graph of the function, as in the DC circuit at the beginning of this section. In effect, the graph of the function “jumps” from one value to another. Someother examples of situations that may be represented by functions with jump discontinuitiesare the population of Ottawa as a function of time, the cost of a taxi ride as a function of distance, and the cost of mailing a first-class letter as a function of its mass. Such functionscan be specified by using different formulas in different parts of their domains.

Remember that a function is a rule. For this particular function, if t [0, 2], the value of v(t ) is 2 2t . If t (2, ), the value of v(t ) is (t 2)2. For instance, we compute v(1), v(2),and v(3) as follows.

Since 1 2, Since 2 2, Since 3 2,v(1) 2 2(1) v(2) 2 2(2) v(3) ((3) 2)2

0 2 1We can graph piecewise functions using a graphing calculator or graphing software.

We now investigate the limiting behaviour of f (x) as x approaches 2 from the left and fromthe right, that is, for x-values less than and greater than 2.

v t t t t t

( ) [ , ]( ) ( , )

2 2 0 22 22

iif

f xx

( ) =

1.8

1.98

1.998

0.01

0.0001

0.000001

f ( x)1.9

1.99

1.999

2.1

2.01

2.001

x

1E−4 means 1.0 × 104

or 0.0001.

Window variables:x [0, 4.7], y [3.1, 6.2]



144 MHR Chapter 3

From the table, we see that v(t ) approaches 2 as t approaches 2 from the left, but v(t ) approaches 0 as t approaches 2 from the right. The notation used to

indicate this is

Such limits are called one-sided limits. Since thefunction approaches different values from the left andthe right, we say that the limit does not exist.

For a limit to exist, the one-sided limits must exist andbe equal.

In general, a left-hand limit is written , which is read as “the limit of f (x)

as x approaches a from the left,” and a right-hand limit is written ,

which is read as “the limit of f (x) as x approaches a from the right.” If the values of f (x) approach L more and more closely as x approaches a more and more closely,

with x a, then . Similarly, if we consider only x a, if the values of

f (x) approach L more and more closely as x approaches a more and more closely,then .

As mentioned above, if a function behaves differently to the left and to the right ofa number a, then does not exist. That is,

If , then does not exist.

If , then . (This is true even if f (a) L.)

Example 1 Limits and Discontinuities

a) The graph of a function f is shown. Use it to findthe following limits, if they exist.

i) ii) iii)

iv) v) vi)

vii) viii) ix)

b) There are two discontinuities on the graph.Identify each as a jump discontinuity, an infinite discontinuity, or a removablediscontinuity.

lim ( )x

f x→ 2

lim ( )x

f x→ +2

lim ( )x

f x→ −2

lim ( )x

f x→ 1

im ( )x

x→

+1lim ( )

xf x

→ −1

lim ( )x

f x→ 0

im ( )x

x→ +0

lim ( )x

f x→ −0

lim ( )x a f x L→ =lim ( ) lim ( )x a x af x L f x→ →− += =

lim ( )x a

f x→

lim ( ) lim ( )x a x a

f x f x→ →− +

≠

lim ( )x a

f x→

lim ( )x a

f x L→ +

=

lim ( )x a

f x L→ −

=

lim ( )x a

f x→ +

im ( )x a

x→ −

lim ( )x

v t → 2

lim ( ) lim ( )x x

v t v t → →− +

= − =2 2

2 0and

1 2 3 5

2

4

0 t

6

8

4

–2

v t ( )

v t =( )2 – 2t if [0, 2]t

( – 2)t 2

if (2, )t {

1–1–2 2 3 4

2

4

0 x

y6




Solution

a) From the graph, the values of f (x) approach 3.5 as x approaches 0 from the left as wellas from the right. Therefore,

i) and ii)

iii) Since the left- and right-hand limits are equal, .

From the graph, the values of f (x) approach 3 as x approaches 1 from the left, but theyapproach 0 as x approaches 1 from the right. Therefore,

iv) and v)

vi) Since the left- and right-hand limits are not equal, does not exist.

From the graph,

vii) and viii)

ix) Since the left- and right-hand limits are equal, , despite the fact that

f (2) 1.

b) The discontinuity at x 1 is a jump discontinuity, because the function “jumps” fromone value to another. The discontinuity at x 2 is a removable discontinuity, since it can be

removed by redefining f (2) 2.

Another function with a jump discontinuity is the Heaviside function, named afterthe electrical engineer Oliver Heaviside (1850–1925). This function describes theelectric current in a circuit that is switched on at a specific time, t 0. For t < 0,the current is off, which is represented by H (t ) 0. For t 0, the current is on,which is represented by H (t ) 1.

The Heaviside function is defined by

Example 2 Limits of the Heaviside Function

Evaluate the following limits of the Heaviside function, if possible.

a) b) c) lim ( )t

H t → 0

lim ( )H t → +0

im ( )t

H t → −0

H t ( )

[ , )

0

0

if ( , 0)

1 if

t

t

2 4– 2–4

2

0 t

H t ( )

lim ( )x

f x→

=2

2

lim ( )x

f x→

=2

2lim ( )x

f x

2

2

lim ( )x

f x→ 1

lim ( )x

f x→ +

=1

0lim ( )x

f x→ −

=1

3

lim ( ) .x

f x→

=0

3 5

lim ( ) .x

f x→ +

=0

3 5im ( ) .x

x→ −

=0

3 5



146 MHR Chapter 3

Solution

a) Since H (t ) 0 for t 0, b) Since H (t ) 1 for t 0,

c) Since does not exist.

Example 3 Limits of an Absolute Value Function

Show that .

Solution

Recall that

Therefore,

and

Since the left- and right-hand limits exist and are equal,

Example 4 Period and Length of a Pendulum

The period of a pendulum, T , in seconds, is given approximately by the equation, where x is the length of the pendulum, in metres.

a) Find . b) Explain what the limit in part a) means.

Solution

a) Notice that the function is defined only for x 0, so the two-sided limit

does not exist. As x approaches approaches 0.

Therefore,

limx

x→ +

=0

2 0

0 0 2, ,x x≥limx

x→ 0

2

f x x( ) = 2

limx

x→ +0

2

T x= 2

imx

x→

=0

0

lim lim ( )x x

x x→ →− −

= −

=0 0

0

im imx x

x x→ →

+ +=

=0 0

0

x x x

x x

if

if

[ , )

( , )

0

0

imx

x→

=0

0

lim ( ) lim ( ), lim ( )t t t

H t H t H t → → →− +

≠0 0 0

im ( ) imt t H t → →+ +==

0 0 11

lim ( ) limt t H t → →− −==

0 0 00

2 4–2–4

2

4

0 x

y

y x= | |




b) As the length of the pendulum gets shorter and shorter, the period also gets shorter andshorter. In other words, the shorter the pendulum, the faster it oscillates.

The result of part b) of Example 4 is true not just for pendulums but for practically alloscillating systems in nature. For instance, earthquakes produce relatively slow (although

powerful!) vibrations, but the tiny quartz crystal in a watch vibrates extremely rapidly.

Example 5 Limits of a Function Defined Piecewise

Consider the function

Graph the function, and use the graph to determine whether and

exist.

Solution

We can graph piecewise functions by including the intervals for each piece in thefunction definition.

We first determine the one-sided limits. From the graph, the values of f (x) approach 3 as xapproaches 0 from the left, but they approach 2 as x approaches 0 from the right.

Therefore, and

The left- and right-hand limits are not equal, so does not exist.

However, the values of f (x) approach 0 as x approaches 2 from the left and from the right.

Therefore, and

The left- and right-hand limits exist and are equal, so im ( ) .x

x→

=2

0

lim ( ) .x

f x→ +

=2

0lim ( )x

f x→ −

=2

0

lim ( )x

f x→ 0

lim ( ) .x

f x→

+

=

0

2im ( )x

x→

−=

0

3

lim ( )x

f x→ 2

lim ( )x

f x→ 0

f x

x x

x x

x x

( )

( , ]

( ) [ , )

3 0

2

2 22

if

if (0, 2)

if





148 MHR Chapter 3

Example 6 Infinite Limits

The magnitude of the repulsive force exerted by an electron at the origin on another

electron located at position x on the x-axis is , provided the force and distance

units are chosen appropriately. Determine if it exists.

Solution

First, we graph the function.From the graph, we can see that theleft- and right-hand limits do not exist.For example, as x approaches 0 from the

right, the function values get larger andlarger, without approaching any specific

y-value. The same is true as x approaches 0 from the left. Thus, does not exist.

Example 6 shows the third type of discontinuity, the infinite discontinuity, as also seenin the investigation. If a function does not have a discontinuity at a certain point, it iscontinuous at that point. A more formal definition follows.

If the limit of a function at a certain value exists, and is equal to the value of the function atthat value, then the function is continuous at that value.

A function f is continuous at a number a if .

The definition of continuity implicitly requires three things:1. f (a) is defined (that is, a is in the domain of f )

2. exists and exist and are equal)

3.

Conversely, if a function f is continuous at a number a,

then . If a function has no discontinuities,

it is called a continuous function. From the definition

of a continuous function, if a function f (x) is continuous

at x a, to determine , all we need to do is

evaluate f (a).

Most physical phenomena, such as the position and velocity of a car, air temperature, andthe height of a tree, are continuous. As we saw earlier in this section, a discontinuousfunction involving electric currents can be written in terms of the Heaviside function, whichhas a jump discontinuity at t 0 because does not exist.lim ( )

t H t

→ 0

lim ( )x a

f x→

lim ( ) ( )x a

f x f a→

=

lim ( ) ( )x a

f x f a→ =

lim ( )x a

f x→ +

(lim ( )x a

f x→ −

lim ( )x a

f x→

lim ( ) ( )x a

f x f a→

=

lim ( )x

f x→ 0

l im ( ),x

f x 0

f x x( ) =1

2

( , ( ))a f a

y f x= ( )

a0

x

y

Window variables:

x [3, 3], y [2, 6]




if n is a positive

integer.

lim ( ) lim ( )x a

n

x a

n

f x f x→ →

[ ] =

The limit of a positive integer power is thepower of the limit.

if the root on the

right side exists.

lim ( ) lim ( )x a

n

x anf x f x

→ →= The limit of a root is the root of the limit,

if the root exists.

if existsan

limx a

n nx a→

=

limx a

n nx a→

=

The limit of a quotient is the quotient of thelimits, if the limit of the denominator is not 0.lim

( )

( )

lim ( )

lim ( ) lim ( ) .

x a

x a

x ax a

f x

g x

f x

g x g x

→

→

→→

= ≠if 0

The limit of a product is the product of thelimits.

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

f x g x f x g x→ → →

[ ] =

The limit of a constant times a function isthe constant times the limit of the function.

im ( ) im ( )x a x a

c x c x→ →

[ ] =

The limit of a difference is the difference of the limits.

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


−[ ] = −

The limit of a sum is the sum of the limits.lim ( ) ( ) lim ( ) lim ( )x a x a x a


+[ ] = +

Using the properties of limits, it can be shown that many familiar functions are continuous.Recall that a polynomial is a function of the form P(x) a

nx

n a

n1x

n1 … a

1x a

0,

where a0, a

1,… , a

nare constant. A rational function is a ratio of polynomials.

a) Every polynomial P is continuous at every number a, that is,

b) Every rational function , where P and Q are polynomials, is continuous at

every number a such that Q(a) 0, that is, , provided Q(a) 0.

For instance, f (t ) t 2 6t 5 is a polynomial, so it is continuous at t 3, and therefore,

In this example, as well as others in this section, we are making use of the limit laws.

The Limit LawsSuppose that the limits and both exist and c is a constant.

Then, we have the following limit laws.

lim ( )x a

g x→

lim ( )x a

f x→

lim( ) ( )

( )

t t t f

→+ + =

= + +=

3

2

2

6 5 3

3 6 3 5

32

lim( )( )

( )( )x a

P xQ x

P aQ a→

=

f x P x

Q x( )

( )( )

=

lim ( ) ( ).x a

P x P a→

=

limx a

c c→

=

limx a

x a→

=



150 MHR Chapter 3

Example 7 Evaluating the Limit of a Function at a Point of Continuity

Evaluate the following limits.

a) b)

Solution

a) Since the function is continuous at x 2, the limit is simply f (2).

That is,

b) Since is continuous at x 3,

Example 8 Limit of a Function at a Removable Discontinuity

Evaluate

Solution

Let

Graph the function.

Note the gap, or hole, in the graph at x 2.

We cannot find the limit by substituting x 2 because f (2) is indeterminate (substituting 2

gives ). Remember that, to determine , we must consider values of x that are

close to a but not equal to a. In this example, x 2, so we can factor the numerator as a

difference of squares.

lim ( )x a

f x→0

f x xx

( ) .= −−

2 42

lim .x

xx→

−−2

2 42

lim ( )x

x x→

+ + = + +

=

3

2 22 1 3 2 3 1

16

= 4

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

lim( )

x

x xx

2

2 23 51

2 3 2 53

33

1

f x x x

x( )

2 3 51

limx

x x

3

2 2 1limx

x xx 2

2

3 51

Window variables:

x [4.7, 4.7], y [3.1, 6.2]

Window variables:

x [4.7, 4.7], y [18.6, 9.3]

Window variables:x [4.7, 4.7], y [0, 6.2]




Notice in Example 8 that we replaced the given rational function by a continuous function, g (x) x 2, that is equal to f (x) for x 2. Notice also that the graph of f (x) above isidentical to the graph of g (x) except at x 2. This is a removable discontinuity.

lim lim ( )( )

( )

lim( )

x x

x

xx

x xx

x

2

2

2

2

42

2 22

2

2 2

4

Key Concepts

The limit of a function is written as , which is read as “the limit of f (x), as x approaches a, equals L.”

If the values of f (x) approach L more and more closely as x approaches a more andmore closely (from either side of a), but x a, then .

The left-hand limit of a function is written as , which is read as “the limit

of f (x) as x approaches a from the left.”

If the values of f (x) approach L more and more closely as x approaches a more andmore closely, with x a, then .

The right-hand limit of a function is written as , which is read as “the

limit of f (x) as x approaches a from the right.”

If the values of f (x) approach L more and more closely as x approaches a more andmore closely, with x a, then .

In order for to exist, the one-sided limits and must both

exist and be equal. That is,

If , then does not exist.

If , then .

To check that a function f (x) is continuous at x a, check that the following threeconditions are satisfied:

a) f (a) is defined (that is, a is in the domain of f (x))

b) exists

c)

Also, if f (x) is continuous at x a, then .

Every polynomial P is continuous at every number, that is, lim ( ) ( ).x a

P x P a→

=

lim ( ) ( )x a

f x f a→

=

lim ( ) ( )x a

f x f a→

=

lim ( )x a

f x→

lim ( )x a

f x L→

=lim ( ) lim ( )x a x a

f x L f x→ →− +

= =

lim ( )x a

f x→

im ( ) im ( )x a x a

x x→ →− +

≠

lim ( )x a f x→ +lim ( )x a f x→ −lim ( )x a f x→

lim ( )x a

f x L→ +

=

lim ( )x a

f x→ +

lim ( )x a

f x L→ −

=

lim ( )x a

f x→ −

lim ( )x a

f x L→

=

lim ( )x a

f x L→

=

We are not dividing by 0 since x ≠ 2.

We can substitute because g ( x ) =

x +

2 is continuous.



152 MHR Chapter 3

Every rational function , where P and Q are polynomials, is continuous

at every number a for which Q(a) 0, that is, , Q(a) 0.

Discontinuities:

a) If a function f (x) has a removable discontinuity at x a, then

exists, and the discontinuity can be removed by (re)defining f (x) L at the singlepoint a.b) If a function has a jump discontinuity, the function “jumps” from one value toanother.

c) If a function f (x) has an infinite discontinuity at x a, the absolute values of the function become larger and larger as x approaches a.

Communicate Your Understanding

1. Suppose that .

a) How is this equation read?b) What does the equation mean?c) Is it possible for the equation to be true if f (3) 4? Explain.d) Is it possible for the equation to be false if f (3) 7? Explain.

2. Suppose that and .

a) What kind of limit is each equation? What does each equation mean?b) Does exist? What kind of discontinuity exists at x 2? Explain.

3. a) Can the two-sided limit ever exist at a jump discontinuity? an infinitediscontinuity? Explain using diagrams.b) For which type of discontinuity does the two-sided limit always exist? Explain.

4.Given that

explain why each of the following limits does not exist.

a) b) c)

5. a) Explain in your own words the conditions that must be met for a function f (x)to be continuous at a number a.b) A student claims, “If the left- and right-hand limits of a function, as x approaches a,

are equal, then the function is continuous at a.” Discuss the validity of this statement.c) Draw examples of each of the following discontinuous functions.

i) f (a) is defined, exists,

ii) f (a) is defined, does not exist

6. Explain, using examples, when substitution can be used to solve a limit.

lim ( )x a

f x→

lim ( ) ( )x a

f x f a→

≠lim ( )x a

f x→

lim ( )( )x

g xf x→

−

3

4lim

( )

( )x

f x

f x 2 4lim

( )

( )x

f x

g x→ 2

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

f x f x g x g x→ → → →

= = = = −2 3 2 3

4 0 0 3

lim ( )x

f x→ 2

lim ( )x

f x→ +

=2

5lim ( )x

f x→ −

=2

1

lim ( )x

f x→

=3

7

lim ( )x a

f x L

lim( )( )

( )( )x a

P xQ x

P aQ a→

=

f x P x

Q x( )

( )( )

=




A

Practise

1. Use the graph of f (x) to state the value of

each limit, if it exists. If it does not exist,explain why.

a) b)

c) d)

e)

2. Use the graph of g (x) to state the value of each limit, if it exists. If it does not exist,explain why.

a) b)

c) d)

e) f )

g) h)

i) j)

k) l)

m) n)

3. Use the graph of g (x) to state the valueof each quantity, if it exists. If it does notexist, explain why.

a) b)

c) d) g (2)

e) f )

g) h) g (5)

i ) j)

4. Identify each discontinuity in questions1 to 3 as removable, jump, or infinite.

5. Let

Sketch the graph of f (x). Then, find each limit,if it exists. If the limit does not exist, explainwhy.

a) b) c)

6. Let

Sketch the graph of g (x). Then, find each limit,if it exists. If the limit does not exist, explainwhy.

a) b) c) lim ( )x

g x→ 2

im ( )x

g x→ +2

lim ( )x

g x→ −2

g x x x

x x( ) ( , ]

( , )

2 2

1 2

if

if

lim ( )x

f x→ 1

lim ( )x

f x→

+1im ( )

xx

→ −1

f x x

x x

( )( , )

[ , )

2 1

3 1

i

if

lim ( )x

g x→ 5

im ( )x

g x→ +5

lim ( )x g x→ −5

im ( )x

g x→ 2

im ( )x

g x→ +2

im ( )x

g x→ −2

lim ( )x

g x→ 0

lim ( )x

g x→ −2

2 4 6 8–

2–

4

4

0

x

y

8

12

–6

16

–4

y g x= ( )

im ( )x

g x→ −8

lim ( )x

g x→ 6

lim ( )x

g x→ +6

lim ( )x

g x→ −6

lim ( )x

g x→ 4

lim ( )x

g x→ +4

im ( )x

g x→ −4

lim ( )x

g x→ 2

lim ( )x

g x→ +2

lim ( )x

g x→ −2

lim ( )x

g x→ 1

lim ( )x

g x→ +1

lim ( )x

g x→ −1

lim ( )x

g x→ − +3

2 4 6 8–2–4

2

4

0 x

y6

y x= ( ) g

lim ( )x

f x→ 6

lim ( )x

f x→ 4

lim ( )x

f x→ 2

im ( )x

x→ 0

lim ( )x

f x→ −1

2 4 6 8–2–4

2

4

0 x

y

–2

y f x= ( )



154 MHR Chapter 3

7. Let

Sketch the graph of h(x). Then, find each limit,if it exists. If the limit does not exist, explainwhy.

a) b) c)

8. Let

Sketch the graph of f (x). Then, find each limit,if it exists. If the limit does not exist, explainwhy.

a) b) c)

9. Using a graphing calculator or graphingsoftware, find each limit, if it exists.

a) b)

c) d)

e) f )

g) h)

10. Determine the value of each limit.

a) b)

c) d)

e) f )

g) h)

i) j)

11. Find each limit.

a) b)

c) d)

e) f )

g) h)

i) j)

k) l)

Apply, Solve, Communicate

12. Communication Use the graphs of f (x) and g (x) to evaluate each limit, if it exists. If thelimit does not exist, explain why.

a) b)

c) d)

e) f )

g) h)

i) j)

y

2 4–2

2

0 6

–2

x

y g x= ( )

2 4 6–2–4

2

4

0 x

y

6

y f x= ( )

lim ( )

( )x

g x

f x→ −7lim

( )

( )x

f x

g x→ −4

lim ( ) ( )x

f x g x→

+[ ]4

lim ( )x

f x→ 4

lim(

( )x

g x

f x→ 0lim

( )( )x

f x

g x→ 0

lim ( ) ( )x f x g x→ −[ ]0lim ( ) ( )x f x g x→ 2

lim ( ) ( )x

f x g x→ −

[ ]2

lim ( ) ( )x

f x g x→ −

+[ ]2

limx

xx→

−

−3

1 133lim

x

x

x→

−

−4

4

2

limx

xx→ −

++2

3 82

limx

x

x→

−

−3 3

3

27

limx

x

x→

−−2

3

2

8

4limx

x x

x x→

− −− +1

2

2

5 3 2

3 7 4

limx x xx x→ + −− +4

2

2 2 246 8limx x xx→ − + +−3

2

22 7 39

limx

x

x x→ −

+

− −2 2

2

3 10limx

x xx→

− +

−4

2 6 84

limx

x

x→

−

−2 2

2

4lim

xx→

−

−3

2 93

limt

t t → −

+2

4 23 4limx

x x

x x→

+ ++3

2

2

8 12

2

limx

xx→

−

+0

42

lim( )x

x x x→

− + −1

3 2 1

lim ( )x

x x→ −

− +1

23 4 10lim( )x

x→

−2

3 4

limx

x→ 4

2imx

x→ 5

mx → 6

πlimx → 2

4

limx

x

x→ 0

limx

x

x→

+

0

limx

xx→ −0

limx

x→

−4

4

limx

x→

+

−4

4limx

x→

−

−4

4

limx

x→

−2

4 4 2limx

x→ +0

4

lim ( )x

f x→ − 1

lim ( )x

f x→ − +1

lim ( )x

f x→ − −1

f xx x

x

x x

( )( , )

( , )

2

11 1

1

ifif

if

lim ( )x

h x 0

lim ( )x

h x→ +0

lim ( )x

h x→ −0

h x

x x

x

x x

( )

( , )

( , )

2 0

0 0

2 0

i

if

if

B





156 MHR Chapter 3

c) Find the following for I (R).i) ii)

iii) I (2) iv)

20. Consider the function f (x) described by

Find each of the following, if it exists.

a) b)

21. Find each limit, if it exists.

a)

b)

c)

22. Find functions f (x) and g (x) such

that exists but and

do not exist.

23. Evaluate

24. Let

Sketch the graph of f (x) and determine thevalues of x at which f (x) is discontinuous.

25. For what values of c is the function

continuous at every number?

f x cx x

c x x( )

( ) ( , )

[ , )

1 2

1 2

3

2 2

if

if

f x

x x

x x x

x x( )

,

, ) ( ,

( ) (

1 1 1

1 2 1 1 2

3 22

if

if or

if ,, )

( ) ( , )

x x

3 22 if

lim .x

x

x→

−−1 3

1

1

lim ( )x

g x→ 0

lim ( )x

f x→ 0

lim ( ) ( )x

f x g x→

+[ ]0

limx

x

x→

− −− −3

7 2

1 4

limx

x

x→

−

−8 3 2

limx

x

x→

−− +3

2

2

3

3 6

lim ( )x

f x→ 2

im ( )x

x→ 1

2

f x x

x( ) =

1

0

if is an integer

if is not an integer

i m ( )R

I R→ 2

im ( )R

I R→ +2

lim ( )R

I R→ −2

Application

Communication

Thinking/Inquiry/Problem Solving

Knowledge/Understanding The greatest integer function is defined by the largest integer

that is less than or equal to x. For instance,

a) Sketch the graph of this function.b) Find and .

c) For what values of a does exist?

d) For what values of x is the greatest integer function discontinuous?e) Sketch the graph of Where is it discontinuous?

f ) Sketch the graph of Where is it discontinuous?h x x x( ) .= −

g x x( ) = +2 1

limx a

x→

limx

x→

+3 im

xx

→ −3

3 4 2 5 , . . an6 6 6 83 6 , . ,

x

Achievement CheckAchievement Check

3-4 the Limit of a Function

Documents