72 Limits and Continuityr

If lim f(x) = L, then by Definition 2.1.1, for any E > 0 there exists a -5:>x-a

such that statement (6) holds. By replacing x by t + a and x - a by t wethe equivalent statement (5). Thus from Definition 2.1.1 we can conclude

limf(t + a) = L,-0

EXERCISES 2.2

In Exercises 1 through 14, find the limit and, when applicable,indicate the limit theorems being used.

In Exercises 21 through 39, find the limit and, when app .indicate the limit theorems being used.

r -2522. lim ---.--5 z+53x -I

24. lim -2--._1/39x - 1.

3x2 -17x + 2026. lim ""-'2'---::-::---:-

._44x - 25x + 36S3 - I

28.lim--._1 s - I

8t3 - 2730. lim

,-3/2 4t2 - 9

32. lim Jh+5 - 2~__ I h + 1

34 I· if;- I.Im---.-1 x-I

. Xl - X2 - X + 1036. lim ---=-----

.--2 x2+3x+2

2 I· X2 - 491. Im-::--

7%-7 .•..•.-

AX2 -923. lim ---

.--3/2 2x + 325 li 352

- 85.0-'16• s~ 2s" - 95 + 4

27. lim y3 + 8~--2 Y + 2

r-r-r-rx-r-r-r-r-r-y2 _ 929. lim

y--3 2j!!-+7y+3.r;. - 131. lim---.-1 x-I

33. lim ~ - .j7:

1. liin (3x - 7).-53. Jim (x2 +-2x - I)

x-2

2. lim (5x + 2)

4. lim (2x2 - 4x + 5)",-3

5. lirn (Z3 + 8)s-- 2

6. lim (y3 - 2y2 + 3y - 4)1'--1

1:_ 4x - 5 8 I' 3x + 47. WL-- . Im--"-3 5x - 1 ",-2 8x - I

9. lim t2

3- 5 10 r 2x + I

,_22t + 6 •• ~I~.\X2 - 3x + 4

11. Jim I~+1 12. lim X2 + 3x + 4,_1 'V r + 3 .-2 x3 + 1

13. I· X2 ;- 3x + 4 . f1!§+ Zxim 3 - 14. lim 3 .--"_42x2--x-l •__ 35-x

In Exercises 15 through 20, do the following: (a) Use a calculatortotobu/ate to four deci1lU!1places values of f(x) for the specified

. oai~s of x. What does f(x) appear to be approaching as x ap-proaches c? (b) Find limf(x) and, when applicable, indicate the

limit theorems being used.

x-215. f(x) = -2--; x is I, 1.5, 1.9, 1.99, 1.999 and x is 3, 2.5, 2.1,

. x-42.01, 2.001; c = 2

2X2 + 3x - 216. f(x) = 2 6 16; x is -3, -2.5, -2.1, -2.01, -2.001x - x- .

and x is -I, -1.5, -1.9, -1.99, -1.999; c = - 2X2 + 5x + 6 .

17. f(x) = 2 ; X IS -4, -3.5, -3.1, -3.01, -3.001,x - x -12

-3.0001 and x is -2, -2.5, -2.9, -2.99, -2.999, -2.9999;c =-3

2x -·318. f(x) = -2--; x is I, 1.4, 1.49, 1.499, 1.4999and x is 2, 1.6,

4x -91.51, 1.501. 1.5001; c = 1

3-';;19. f(x) = 9 _ x; x is 8. 8.5, 8.9, 8.99, 8.999 and x is 10, 9.5,

9.1,9.01, 9.001; c = 92-J4-x .lO. f(x) = ;x IS -1, -0.5, -0.1, -0.Q1, -0.001 and

xx is 1, O.S,0.1, 0.01, 0.001; c = 0

35. lim {fh+l - 1~-o h'

2X2 - X - 337. lim 3 2 ~

._ - I X + 2x + 6x + 5

I. 2y3 - IIy2 + 10y + 838. im 3 2

y~4 3y - 17y + 16y + 162x3 - 5x2 - 2x - 3

39. Jim 3 2._34x - I3x + 4x - 3

40. If f(x) = X2 + 5x - 3, show that lim f(x) = f(2) .•-2

41. If F(x) = 2x3 + Tx - I, show that lim F(x) = F( -I~%--1

X2 - 1642. If g(x) = ---, show that lim g(x) = 8 but that g(4

x - 4 .-4not defined.'

-Jx + Y - J43. If h(x) = , show that 1im h(x) = t, but thatx %-0 .

is not defined.44. Given that f is the function defined by

_ {2X - 1 if x*-2fix) - 1 if x = 2

(a) Find Jim f(x), and show that lim fix) * f(2). (b) Dra%-2 %-2

sketch of the graph of f.

-

tsalJ>O

t we havelclude that

•=====

255- 1-:1..17x + 20~11

~-;ftC905-2h+l; -1-=1

3 X2 - X + 10~

X2 ~ 3x + 2

f(X) 0= f(2).

im F(x) 0= F(-1)._-I

III h(x) 0= t, but t-0

:d by

2.3 One·Sided Limits

4S. Given that f is the function defined by

(;.:2-9 ifx#-3f(x) = t4 if x = - 3

(a) Find lim f(x), and show that lim f(x) # f( - 3). (b)%--3 x--3

Draw a sketch of the graph of f.Definition 2.1.1 to prove that if

/(x) = Land lim g(x) = M

then

Iim [f(x) - g(x)] = L - M

Prove Limit Theorem 5 by applying Limit Theorem 4 andematical induction.

'fe Limit Theorem 7 by applying Limit Theorem 6 and:::alhc:matical induction.

Definition 2.1.1 to prove that if

f(x) = Land lim g(x) = 0

then

lim (f(x) . g(x)] = 0

(Hint: To proi ; that lim (f(x) . g(x)] = 0 we must show that%-.for any" > 0 there exists a .5 > 0 such that if 0 < [x - al < .5,then If(x," g(x)1 < E. First show that there is a .5. > 0 suchthat if 0 < [x '- al < .5., then If(x) I < 1 + ILl, by applyir..gDefinition 2.1.1 to lim f(x) = L with" = 1 ·and .5= .5., and

then use the triangle inequality. Then show that there is a.52> 0 such that if 0 < Ix - al < .52, then 19(x)1< E/(1 + ILl).by applying Definition 2.1.1 to lim g(x) = O. By taking .5 as

the smaller of the two numbers .5. and .52, the theorem isproved.)

SO. Prove Limit Theorem 6: If Jim f(x) = Land lim g(x) = M,

then

lim (f(x)· g(x)] = L . M

(Hint: Let f(x) . g(x) = (f(x) - L Jg(x) + L[g(x) - M] + L . M.Apply Limit Theorem 5 and the result of Exercise 49.)

.3 ONE-SIDED LIMITS When considering Jim f(x), we are concerned with values of x in an open interval

containing Q but not at Q itself, that is, at values of x close to Q and eithergreater than Q or less than Q. However, suppose that wehave the function ffor which, sav, f(x) = -Ix - 4. Because f(x) does not exist if x < 4, f is notdefined on any open interval containing 4. Hence lim -Ix - 4 has no meaning.

%-4However, if x is restricted to values greater than 4, the value of -Ix - 4 can bemade as close-to 0 as we please by taking x sufficiently c'ose to 4 but greaterthan 4. In such a case we let x approach 4 from the right and consider the one-sided limit from the right, O{ the right-hand limit, which is now defined.

2.3.1 DEFINITION Letf be a function that is defined at every number in some open interval (a, c).Then the limit off(x), as x approacbes a from the right, is L, written

lim f(x) =Lx--+a

if for any £ > 0, however small, there exists a ~ > ° such that

if O<x-a< s then If(x)-LI<£

Note that in the preceding statement there are "no absolute-value bars aroundx - a since the condition x > a is equivalent to x - a > O.

It follows from Definition 2.3.1 that

Jim -Ix - 4 = 0

If when considering the limit of a function the independent variable x isrestricted to values less than a number a, we say that x approaches a from theleft; the limit is called the one-sided limit from the left, or the left-hand limit.

73

7. Lilllits and Contjnuity

EXAMPLE 4 Let f be defined by

{

X + 5 if x < -3f(x) = J9 - X2 if - 3 :$; x :$; 3

3-x if3<x

(a) Draw a sketch of the graph of f. (b) Fino, if they exist, each of the followinglimits: Jim f(:x.), lim f(x), lim f(x), lim f(x), Jim f(x), Jim f(x).

%--3- %--3+ x--3 x-3- %-3+ %-3

Solution(a) A sketch of the graph of f is shown in Figure 5.

(b). lim f(x) = lim (x + 5) lim .f(x) = lim ~%--]- %--3- %~-3+ %--]+

y=2 =0

Because lim f(x) # lim f(x), then lim f(x) does not existx--3- %--3+ %--3

lim f(x) = lim ../9- :-;2 lim f(x) = lim (3 -- x)%-3- %-3- %-3+ %-3·

=c =0Because Jim ((x) = lim f(x), then Jim f(x) exists and IS O.

%-3- x-3+ %-3AGURE 5

EXERCISES 2.3

In Exercises I through 22, draw a sketch of the graph, and findthe indicated limit if it exists; if the limit does not exist, 'state thereason.

6.h{2X+! ifx<3

(x) =10 - x if3 ~ x

(a) limh(x}; (b) lim h(x}; (c) lim h(x}%-3+ %-3- %-3

{

2 ifx<!1. f(x) = - 1 if x = !

-3 ifl<x

(a) lim f(x}; (b) lim f(x); (c) Jim f(x}",-1+ %-1 "'-J

{

2r+3 if}<l

7. g(r} = 2 if r = 17-2r ifl<r

(a) lim g(r}; (b) Jim g(r); (c) lim g(r}r-1+ ,-1- ,-1

{-2 if x -c O

1. f(x} = 2 if 0 ~ x

(a) lim f(x}; (b) lim f(x}; (c) Jim f(x)%-0" .x-O - x-o

{

3+ t2 if t < -28. g(t) = 0 if t = - 2

11 - t2 if - 2 < t(a) Jim g(t}; (b) lim g(t}; (c) lim g(t)

'--2+ '--2- 1--2{t + 4 ift ~ -4

3. f(t} = 4 _ t if -4 < r

(a) lim f(r); (b) Jim f(r}; (e) Jim f(r)\--4.:'" \--,,- \--4.

~

2_4 ifx<2

9. ((x\ = 4 if x = 24_X2 if2<x

(a) lim f(x~ (b) lim f(x); (c) lim f(x}%-2· %-2- %-2

15+3 ifss-24. g(s) = 3 if 2-s - < s

(a) Jim g(s); (b) Jim g(s}; (e) Jim g(s}.--2+ . .1--2- s--2

{

X2 if x s 2S. F(x} = 8 _ 2x if 2 < x

(a) lim F(x~ (b) lirn ·F(x}; (c) lim F(x):.:-2" x-2- %-2

{

2x + 3 if x < 110. f(x} = 4 if x = 1

X2 + 2 if 1 < x

(a) lim f(x~ (b) Jim f(x}; (e) lim f(x}%-1'" %-1 x-I

Exen:i_2.3

F{x) = Ix- 51Jim F(x~ (b) lim F(x~ (e) lim F(x)

x~S· x~5 %-5

)= 3 + Ill: - 41lim f(x~ (b) lim fIx); (e) lim fIx)

r-l+ %-2- x-l

)=12x-31-4lim G(x~ (b) lim G(x~ (e) lim G(x)

~3/2· % •...•3/2- %-3/2

{Ix - II if x < -I

F{x) = 0 if x = -1

11- xl if -1 < xlim F(x~ (b) Jim F(x); (e) lim F(x)

.---1+ %--1- %--1

/(x) =.ELx

lim f(x); (b) lim fIx); (e) lim fIx)x~o· %-0- x-O

S(x) = [sgn xl (sgn x is defined in Example I)<a) limS(x); (b) lim S(~); (e) lim SIx)

x-O· x-o- x-o

=

{

2 if x < -2fIx) = .j4 - X2 if -2 ~ x ~ 2

-2 if2<xf

(a) Jim fIx); (b) Jim fIx); (c) Jim fIx); (d) Jim fIx);",,--2- .%--2+ x .....•-2 x-2-

(e) lim f(x~ (f) Jim f(x)%-2· %-2

{

X+l ifx<-If\X) = xl if - 1 ~ x ~ I

2-x ifl<x(a) Jim fIx); (b) Jim fIx); (c): Jim fIx); (d) lim fix);

.x- - 1 - x- - 1 • x- - 1 x-I - ~

(e) Jim fIx); (f) lim fIx)x-l+ x-I

{if, ift < 0

fIt) = Ji if 0 ~ t

(a) Jim fIt); (b) lim fIt); (e) Jim fIt),-0· ,-0- ,-0

{h ifx s 0

g(x) = {/x if0 < x

(a) lim g(x); (b) Jim g(x~ (e) Jim g(x)x-O· %-0- x-o

{

.jx2 - 9 if x s -3F(x) = .j9 - Xl if - 3 < x < 3

. .jxl - 9 if 3 ~ x

(a) lim F(x); (b) Jim F(x); (e) lim F(x); (d) lim F(x);.•.--3- %--3+ %--3 x-3-

(e) Jim F(x); (f) lim F(x),,-3+ %-3

{rr+! if Is - I

G(t) = ,Jl=? if -I < t < I

Fl if I~t

77

(a) Jim G(t); (b) lim G(t); (c) lim GII); (d) Jim G(I);'--1- .--1· .--1 .-1-

(e) lim G(t); (f) Jim G(t)'-1 + '-1

23. F(x) = x - 2 sgn x, where sgn x is defined in Example 1.Find, if they exist: (a) lim F(x); (b) lim F(x); (c) lim F(x).

x-o" x-o - x-o24. h(x) = sgn x - U(x), where sgn x is defined in Example I,

and U is the unit step function defined by

{o ifx<O

U(x) = 1 if 0 ~ x -

Find, if they exist: (a) Jim h(x); (b) Jim h(x); (e) Jim h(x~x~o· %-0- x-O

25. Find, if they exist: (a) Jim [xl; (b) Jim [xl; (c) Jim [x).%-2+ .-2- x-l

26. Find, if they exist: (a) lim [x - 3]; (b) lim [x - 3];x-.· %-4-(c) Jim [x - 3].

x-4

27. Let h(x) = (x - 1) sgn x. Draw a sketch of the graph of h.Find, if they exist: (a) Jim h(x); (b) Jim h(x~ (e) lim h(x).

x-O· %-0- x-O

28. Let G(x) = [xl + [4 - x]. Draw a sketch of the graph of G.Find, if they exist: (a) Jim G(x);-(b) lim G(x); (c) lim G(x).

%-3+ %-3- %-3

_ {3X + 2 if x < 4}29. GIven fIx) = if . Find the value of k such5x + k I 4 ~ x

that lim fIx) exists.x-4

{kx - 3

30. Given fIx) = 2X + k

that lim fix) exists.%--1

ifX<-I}'f -I . Find the value of k suchI - <x

{

Xl if x < -2 }31. Given fIx) = ax + b if -2 < ;;.;< 2 . Find the values of

2x - 6 if2 ~ xa and b such that lim fix) and lim fix) both exist.

.x ....•-2 %-2

{

2X - a if x < - 3 }32. Given fix) = ax + 2b if - 3 ~ x ~ 3 . Find the values of

. b - 5x if3 < xa and b such that lim fix) and lim fIx) both exist.

x--3 %-3

33. Let fIx) = {- 1 ~ffOX < O}. Show that Jim fIx) does not1 I < X x-o

exist but that Jim If(x)1 does exist.x-o

34. r ....r",.. -;-~~cOlcrn2.3.3.35. Shipping charges are often based on a formula that offers a

lower charge per pound as the size of the shipment is in-':It...sed. Suppose x pound. is the weight of a shipment, C(x)dollars is the total cost of the shipment, and

{o.sox if 0 < x s SO

C(x) = 0.70x if 50 < x s 2000.65x if 200 < x

(a) Draw a sketch of the graph of C. Find each of the follow-ing limits: (b) lim C(x); (c) Iim C(x); (d) lim· C(x);

%-50- x-50· %-200-(e) lim C(x). .

x-200·

(b) Show that lim g(x) and lim g(x) both exist but are not%-1- x-l+ .

equal, and hence lim g(x) does not exist..-1

78 Limits and Continuity

. {X2+3 ifx:o;l {X2 ifx<l36. GIven f(x) = . . and g(x) = .-x+l if l x x 2 ifl<x

(a) Show that Iim f(x) and lim f(x) both exist but are not. %-1- x-l+

equal, and hence lim f(x) does not exist..-1

(c) Find formulas for f(x) . g(x).

(d) P.O\e that lim [I(x) . g(x)] exists by showing.-1lim [I(x)' g(x)] = lim [I(x)' g(x)].

.::r-1- x-I·

x 3f(x) = (x _ 2)2

2.4 INFINITE LIMITS In this section we discuss functions whose values increase or decrease withoutbound as the independent variable gets closer and closer to a fixed number.First let us consider the function defined by

3f(x) = (x _ 2)2

The domain of f is the set of all real numbers except 2 and the range is theof all positive numbers. We investigate the function values of f when x is clto 2. Let x approach 2 from the right; in particular, we let xbe 3, 2.5, 2.25, 22.01, and 2.001 and use a calculator to compute the corresponding valuesf(x) shown in Table 1. From this table you see intuitively that as x gets cloand closer to 2 through values greater than 2, f(x) increases without bound.other words, we can make f(x) greater than any preassigned positive num(i.e., f(x) can be made as large as we please) for all values of x close enough2 and x greater than 2.

To indicate that f(x) increases without bound as x approaches 2 throuvalues greater than 2 we write

31248

30030,000

3,000,000

Tallie-I

325225212012001

Table 2

x

31248

30030,000

3,000,000

li 3m ---2 = +:0

.-2+ (x - 2)

Now let x approach 2 from the left; in particular, we let x take on the val1, 1.5, 1.75, 1.9, 1.99, and 1.999 and use a calculator to compute the correspoing values of f(x) appearing in Table 2. You see intuitively from this tableas x gets closer and closer to 2 through values less than 2, f(x) increases withbound; so we write

3f(x)=-·-

(x - 2)211.51.751.91.991.999

v r 3.x~~- (x _ 2)2 = +00

Therefore, as x approaches 2 from either the right or the left,f(x) increases wiout bound, and we write

li 3m ---2 = +00.-2 (x - 2)

~O~~~~--~~I

AGURE 1

From the information in Tables 1 and 2 we obtain the sketch of theof f shown in Figure 1. Observe that both "branches" of the curve get cland closer to the dashed line x = 2 as x increases without bound. This dline is called a vertical asymptote, defined later in this section. .

We now state the formal definition of function values increasing witlwut