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INFINITE LIMITS, LIMITS AT INFINITY, ANDLIMIT RULES
Sections 2.2, 2.4 & 2.5September 5, 2013
INFINITE LIMITS
Consider the function f (x) =1x
.
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
-6-5-4-3-2-1
123456
As x 0+ the value off (x) = 1x grows withoutbound (i.e. 1x ).
As x 0 the value of 1xgrows negatively withoutbound (i.e. 1x ).
By definition a one or two-sided limit must be a real number(i.e. finite), so
limx0+
1x
and limx0
1x
.
do not exist.
INFINITE LIMITS
However, it is convenient to describe the behavior of f (x) = 1xfrom the left or right using limit notation.
So we will write
limx0+
1x
= and limx0
1x
= .
This does NOT mean that the limit is , it is just a descriptionof the behavior of our function f (x) near 0.
We would say:
f (x) = 1x approaches as x approaches 0 from the right.f (x) = 1x approaches as x approaches 0 from the left.
INFINITE LIMITS
It may be the case that both one-sided limits are infinite andalso the same.
For example,
limx0+
1x2
= = limx0
1x2
.
In this case we will again abuse the notation and write
limx0
1x2
= .We would say:
f (x) = 1x2 approaches as x approaches 0.
EXAMPLES
f (x) = sec(x) =1
cos(x)
-5 -4 -3 -2 -1 0 1 2 3 4 5
g(x) =1
(x 7)2
-2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
-4-3-2-1
12345678
(1) limxpi2
f (x) =
(2) limxpi2 +
f (x) =
(3) limxpi2
f (x) = DNE
(4) limx7
g(x) =
(5) limx7+
g(x) =
(6) limx7
g(x) =
VERTICAL ASYMPTOTES
Back to the function f (x) =1x
.
As noted before, theone-sided limits of f at 0are both infinite.
Consider the line x = 0.
The graph of f getsarbitrarily arbitrarily closeto the line x = 0, but it willnever touch it.
We call the line x = 0 a vertical asymptote of the functionf (x) = 1x .
VERTICAL ASYMPTOTES
DefinitionA vertical line x = c is a vertical asymptote of the graph of afunction f if either
limxc+
f (x) = or limxc
f (x) = .
VERTICAL ASYMPTOTES
How do we find vertical asymptotes?
Check wherever a denominator is zero.
For example,
f (x) =1
x 3lim
x3f (x) =
limx3+
f (x) =
Vert. Asym.:x = 3
f (x) =x2 1x + 1
limx1
f (x) = 2lim
x1+f (x) = 2
Vert. Asym.:None
f (x) = tan(x)
For all odd integersn:
limx npi2
f (x) =
limx npi2 +
f (x) =
Vert. Asym.:x = npi/2for all odd n
PRACTICE PROBLEMS
PRACTICE PROBLEMS
Determine the following limits:
(1) limx0
4x2/5
(2) lim(pi2 )
sec() (3) limx0+
(x2
2 1
x
)
Determine the equations of the asymptotes of the followingfunctions:
(1) f (x) =x2 12x + 4
(2) f (x) = sec(x) (3) f (x) =x3 + 1
x2
LIMIT RULES &THE SANDWICH THEOREM
LIMIT RULES
Now that were (hopefully) familiar with limits, it would benice if we had some rules for computing them more quickly.
Let L, M c, and k be real numbers and
limxc f (x) = L and limxc g(x) = M
Sum & Difference Rules
limxc (f (x) g(x)) = LM
Product Rule
limxc (f (x) g(x)) = L M
LIMIT RULES
Let L, M c, and k be real numbers and
limxc f (x) = L and limxc g(x) = M
Quotient Rule
limxc
f (x)g(x)
=LM
if M 6= 0
Power RuleIf r and s are integers with no common factor and s 6= 0
limxc (f (x))
r/s = Lr/s
If s is even, we also need L > 0.
LIMITS OF POLYNOMIALS
For certain functions (continuous) finding the limit is easy.Polynomials are one of those types of functions.
Polynomial Limit RuleLet P(x) be any polynomial, then
limxc P(x) = P(c)
By the quotient rule, if P(x) and Q(x) are polynomial withQ(c) 6= 0, we have
limxc
P(x)Q(x)
=P(c)Q(c)
THE ALMOST THE SAME RULE
The following rule is very useful.
The Almost the Same Rule
If f (x) = g(x) for all x 6= c in some interval open containing c,then
limxc f (x) = limxc g(x)
Its not clear how one would use this, so lets look at anexample.
THE ALMOST THE SAME RULE
Determine the following limit:
limt1
t2 + 3t + 2t2 t 2
Since (1)2 (1) 2 = 0, we cannot use the quotient rule.
However, notice that 1 is also a zero of the numerator:(1)2 + 3(1) + 2 = 0.
This means that (t + 1) is a factor of both the numerator and thedenominator.
limt1
t2 + 3t + 2t2 t 2 = limt1
(t + 1)(t + 2)(t + 1)(t 2) = limt1
(t + 2)(t 2) =
13
THE SANDWICH THEOREM
Sometimes the limit rules arent enough to determine the limitof a function at a given point.
The following theorem allows us to determine the limit of afunction at such a point by sandwiching it between two otherfunctions.
The Sandwich TheoremLet f , g, and h be functions and c and L real numbers. If
limxc g(x) = L = limxc h(x)
andg(x) f (x) h(x)
for all x in some open interval containing c, then
limxc f (x) = L.
THE SANDWICH THEOREM
This is a bit easier to see graphically:
66 Chapter 2: Limits and Continuity
aid of some geometry applied to the problem (see the proof of Theorem 7 in Section 2.4),or through methods of calculus (illustrated in Section 4.5). The next theorem is alsouseful.
The Sandwich Theorem
The following theorem enables us to calculate a variety of limits. It is called the SandwichTheorem because it refers to a function whose values are sandwiched between the valuesof two other functions g and h that have the same limit L at a point c. Being trapped be-tween the values of two functions that approach L, the values of must also approach L(Figure 2.12). You will find a proof in Appendix 5.
The Sandwich Theorem is also called the Squeeze Theorem or the Pinching Theorem.
EXAMPLE 10 Given that
find no matter how complicated u is.
Solution Since
the Sandwich Theorem implies that (Figure 2.13).
EXAMPLE 11 The Sandwich Theorem helps us establish several important limit rules:
(a) (b)
(c) For any function , implies .
Solution
(a) In Section 1.3 we established that for all (see Figure 2.14a).Since we have
(b) From Section 1.3, for all (see Figure 2.14b), and we haveor
(c) Since and and have limit 0 as itfollows that .limx:c (x) = 0
x: c, sxd - sxd - sxd sxd sxd
limu:0 cos u = 1.
limu:0 s1 - cos ud = 0u0 1 - cos u u
limu:0 sin u = 0.
limu:0 u = 0,limu:0 s- u d =u- u sin u u
limx:c (x) = 0limx:c (x) = 0
limu:0 cos u = 1limu:0 sin u = 0
limx:0 usxd = 1
limx:0s1 - sx
2>4dd = 1 and limx:0s1 + sx
2>2dd = 1,limx:0 usxd ,
1 - x2
4 usxd 1 +x2
2 for all x Z 0,
x
y
0
L
c
hf
g
FIGURE 2.12 The graph of issandwiched between the graphs of g and h.
x
y
0 11
2
1
y ! 1 " x2
2
y ! 1 # x2
4
y ! u(x)
FIGURE 2.13 Any function u(x) whosegraph lies in the region between
and haslimit 1 as (Example 10).x: 0
y = 1 - sx2>4dy = 1 + sx2>2d
THEOREM 4The Sandwich Theorem Suppose that forall x in some open interval containing c, except possibly at itself. Supposealso that
Then limx:c sxd = L .
limx:c g sxd = limx:c hsxd = L .
x = cg sxd sxd hsxd
y ! !
y ! !
y ! sin !
!
1
1
" "
y
(a)
y ! !
y ! 1 # cos !
!
y
(b)
2
2
1
112 0
FIGURE 2.14 The Sandwich Theoremconfirms the limits in Example 11.
EXAMPLES
We can use the sandwich theorem to prove two importantlimits:
limx0
sin(x) = 0 and limx0
cos(x) = 1
In order to apply the sandwich theorem we need to find tofunctions to serve as the bread.
In Chapter 1 the following two inequalities are proven(you wouldnt be expected to know these):
|x| sin(x) |x| and 1 |x| cos(x) 1.
So, by the sandwich theorem we have
limx0|x| = 0 = lim
x0|x| = lim
x0sin(x) = 0
limx0
1 |x| = 1 = limx0
1 = limx0
cos(x) = 1.
EXAMPLES
Use the sandwich theorem to determine the following limit:
limx0
x2 sin(x)
(Note that we could find this limit using the product rule.)
In order to apply the sandwich theorem we need to find tofunctions to serve as the bread.
Since 1 sin(x) 1 for all x 6= 0, we know thatx2 x2 sin(x) x2
So, by the sandwich theorem we have
limx0x2 = 0 = lim
x0x2 = lim
x0x2 sin(x) = 0
EXAMPLES
This is a bit easier to see graphically:
PRACTICE PROBLEMS
PRACTICE PROBLEMS
Determine the following limits:
(1) limx2
x3 2x2 + 4x + 8
(2) limt6
8(t 5)(t 7)
(3) limy2
y + 2y2 + 5y + 6
(4) It can be shown that 1 x2
6 x sin(x)
2 2 cos(x) 1.Use this fact to determine
limx0
x sin(x)2 2 cos(x)
LIMITS AT INFINITY
LIMITS AT INFINITY
Up to this point weve only considered the limit of a function ata real number c, that is, as x c.We can also consider the limits of functions as x growspositively or negatively without bound, i.e.
limx f (x) and limx f (x)
Limits at Infinity (intuitive)We say that a real number L is the limit of a function f as xapproaches (resp., ), if the values of f (x) can be madearbitrary close to L by taking x to be sufficiently large (resp.,sufficiently negative).
EXAMPLES
Conisder the function f (x) = 1x again.
61. A function discontinuous at every point
a. Use the fact that every nonempty interval of real numberscontains both rational and irrational numbers to show that thefunction
is discontinuous at every point.
b. Is right-continuous or left-continuous at any point?
62. If functions (x) and g(x) are continuous for couldpossibly be discontinuous at a point of [0, 1]? Give rea-
sons for your answer.
63. If the product function is continuous at must (x) and g(x) be continuous at Give reasons for youranswer.
64. Discontinuous composite of continuous functions Give an ex-ample of functions and g, both continuous at for whichthe composite is discontinuous at Does this contra-dict Theorem 9? Give reasons for your answer.
65. Never-zero continuous functions Is it true that a continuousfunction that is never zero on an interval never changes sign onthat interval? Give reasons for your answer.
66. Stretching a rubber band Is it true that if you stretch a rubberband by moving one end to the right and the other to the left,some point of the band will end up in its original position? Givereasons for your answer.
67. A fixed point theorem Suppose that a function is continuouson the closed interval [0, 1] and that for every x in[0, 1]. Show that there must exist a number c in [0, 1] such that
(c is called a fixed point of ).scd = c
0 sxd 1
x = 0. ! gx = 0,
x = 0?x = 0,hsxd = sxd # g sxd
(x)>g (x) 0 x 1,sxd = e1, if x is rational
0, if x is irrational
2.6 Limits Involving Infinity; Asymptotes of Graphs 97
68. The sign-preserving property of continuous functions Let be defined on an interval (a, b) and suppose that at somec where is continuous. Show that there is an interval
about c where has the same sign as (c).
69. Prove that is continuous at c if and only if
70. Use Exercise 69 together with the identities
to prove that both and are continuousat every point
Solving Equations GraphicallyUse the Intermediate Value Theorem in Exercises 7178 to prove thateach equation has a solution. Then use a graphing calculator or com-puter grapher to solve the equations.
71.
72.
73.
74.
75.
76.
77. Make sure you are using radian mode.
78. Make sure you are using radianmode.2 sin x = x sthree rootsd .cos x = x sone rootd .x3 - 15x + 1 = 0 sthree rootsd2x + 21 + x = 4xx = 2xsx - 1d2 = 1 sone rootd2x3 - 2x2 - 2x + 1 = 0x3 - 3x - 1 = 0
x = c .g sxd = cos xsxd = sin x
sin h sin ccos h cos c -cos sh + cd =cos h sin c ,sin h cos c +sin sh + cd =
limh:0 sc + hd = scd .
sc - d, c + dd
scd Z 0
2.6 Limits Involving Infinity; Asymptotes of GraphsIn this section we investigate the behavior of a function when the magnitude of the inde-pendent variable x becomes increasingly large, or . We further extend the conceptof limit to infinite limits, which are not limits as before, but rather a new use of the termlimit. Infinite limits provide useful symbols and language for describing the behavior offunctions whose values become arbitrarily large in magnitude. We use these limit ideas toanalyze the graphs of functions having horizontal or vertical asymptotes.
Finite Limits as
The symbol for infinity does not represent a real number. We use to describe thebehavior of a function when the values in its domain or range outgrow all finite bounds.For example, the function is defined for all (Figure 2.49). When x ispositive and becomes increasingly large, becomes increasingly small. When x isnegative and its magnitude becomes increasingly large, again becomes small. Wesummarize these observations by saying that has limit 0 as or
or that 0 is a limit of at infinity and negative infinity. Here are precise definitions.
sxd = 1>xx: -q , x: qsxd = 1>x1>x1>x x Z 0sxd = 1>x
qsq d
x:
x: ;q
T
y
0
1
111 2 3 4
2
3
4
x
1xy !
FIGURE 2.49 The graph of approaches 0 as or .x: -qx: q
y = 1>x
Notice that as x, thevalue of f (x) 0. So,
limx
1x
= 0
Similarly, as x , thevalue of f (x) 0. So,
limx
1x
= 0
EXAMPLES
For any integer n > 0, the graph of f (x) =1xn
is given by:
n evenn odd
So we see that
limx
1xn
= 0
for all integers n 1.
EXAMPLES
The limit of a function at need not exist:
f (x) = cos(x)
limx cos(x) = DNE
f (x) = x3 x
limx x
3 x =
limx x
3 x =
f (x) = ex
limx e
x =lim
x ex = 0
LIMIT RULES
The limit rules also hold for limits at :
Limit RulesLet L and M be real numbers and suppose
limx f (x) = L and limx g(x) = M
Then:lim
x(f (x) g(x)) = LM
limx(f (x) g(x)) = L M
limx(f (x)/g(x)) = L/M if M 6= 0
limx(f (x))
r/s = Lr/s
(r, s are integers with no common factors, and L > 0 if s is even)
THE SANDWICH THEOREM
The sandwich theorem also hold for limits at :
The Sandwich Theorem (for limits at )Let f , g, and h be functions and L real numbers, with
limx g(x) = L = limx h(x)
andg(x) f (x) h(x)
for all sufficiently large x, then
limx f (x) = L.
(The analogous statement holds for limits at .)
EXAMPLE
Use the sandwich theorem to determine the following limit
limx
sin(x)x
We need to find functions that we can sandwich sin(x)/xbetween for sufficiently large and sufficiently negative x.
Since 1 sin(x) 1, for all x 6= 0 we have
1x sin(x)
x 1
xSo, by the sandwich theorem
limx
sin(x)x
= limx
1x
= 0.
EXAMPLE
Graphically, we have
LIMITS OF RATIONAL FUNCTIONS
Even with our limit rules, determining limits at can bedifficult.
However, for rational functions finding these limits is relativelyeasy.
For example, suppose f (x) =x 1
4x2 + 2and we want to
determine limx f (x).
First, we rewrite f (x) by dividing the numerator and thedenominator by the term involving the largest power of x. In thiscase the term 4x2:
limx
x 14x2 + 2
= limx
x4x2 14x21 + 12x2
.
LIMITS OF RATIONAL FUNCTIONS
Next we can apply the limit rules, specifically the quotient rule:
limx
x 14x2 + 2
= limx
x4x2 14x21 + 12x2
=lim
xx
4x2 lim
x1
4x2
limx 1 + limx
12x2
=0 01 + 0
= 0
Notice: The degree of the denominator was greater than that ofthe numerator and the limit was 0.
LIMITS OF RATIONAL FUNCTIONS
Suppose f (x) =x2 1
4x2 + 2.
We rewrite f (x) by dividing the numerator and thedenominator by the term involving the largest power of x andemploy the quotient rule:
limx
x2 14x2 + 2
=lim
x14 lim
x1
4x2
limx 1 + limx
12x2
=14 01 + 0
=14
Notice: The degree of the denominator was equal to that of thenumerator and the limit was a non-zero constant.
LIMITS OF RATIONAL FUNCTIONS
Suppose f (x) =x3 1
4x2 + 2.
We rewrite f (x) by dividing the numerator and thedenominator by the term involving the largest power of x anduse the quotient rule:
limx
x3 14x2 + 2
=lim
x 1 limx1x3
limx
4x+ lim
x2x3
=1
limx
4x+ lim
x2x3
=
Notice: The degree of the denominator was less than that of thenumerator and the limit was infinite.
LIMITS OF RATIONAL FUNCTIONS
More generally we have:
TheoremIf f is a rational function, with
f (x) =anxn + + a1x + a0bmxm + + b1x + b0 .
Thenlim
x f (x) = limxanxn
bmxm.
Note that if m n
limx f (x) = limx
anxn
bmxm=
{0 m > n
anbm m = n
EXAMPLES
Determine the following limits:
limx
7x3
x3 3x2 + 6x = limx7x3
x3= lim
x 7 = 7
limx
3x + 7x2 2 = limx
3xx2
= limx
3x
= 0
limx
2x + x2
x 4 = limxx2
x= lim
x x =
limx
x4x2 + 7x4 7x2 + 9 = limx
x47x4
= limx
17
= 17
HORIZONTAL ASYMPTOTES
DefinitionWe say that a function f has a horizontal asymptote of y = L ifeither
limx f (x) = L or limx f (x) = L.
For example, as was just shown limx
7x3
x3 3x2 + 6x = 7.
So f (x) = 7x3
x33x2+6x has ahorizontal asymptote ofy = 7.
-1
01234567891011121314
PRACTICE PROBLEMS
PRACTICE PROBLEMS
For each of the functions below, determine the limit as x and the equation of any horizontal asymptotes:
(1) f (x) =3 (2/x)4 + (1/x2)
(2) g() =cos()
(3) h(y) =y + 2
y + 5 + 6y2
(4) q(x) =7 + 3x2
x + 6
Infinite LimitsInfinite LimitsInfinite LimitsExamplesVertical AsymptotesVertical AsymptotesVertical AsymptotesPractice ProblemsLimit RulesLimit RulesLimits of PolynomialsThe ``Almost the Same'' RuleThe ``Almost the Same'' RuleThe Sandwich TheoremThe Sandwich TheoremExamplesExamplesExamplesPractice ProblemsLimits at InfinityExamplesExamplesExamplesLimit RulesThe Sandwich TheoremExampleExampleLimits of Rational FunctionsLimits of Rational FunctionsLimits of Rational FunctionsLimits of Rational FunctionsLimits of Rational FunctionsExamplesHorizontal AsymptotesPractice Problems