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Sep 24, 2015

math limits

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)

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