2.3 Calculating Limits Using the Limit Laws

[Pages:3]2.3 Calculating Limits Using the Limit Laws

Calculating limits by testing values of x close to a is tedious. The following Theorem essentially says that any `nice' combination of functions has exactly the limit you'd expect.

Theorem.

Suppose

lim

xa

f (x) and

lim

xa

g(x)

both

exist

and

that

c

is

constant.

Then the following limits exist and may be computed.

1. xlima c = c

2.

lim

xa

x

=

a

3. xlima c f (x) = c xlima f (x)

4.

xlima[

f

(x)

?

g(x)]

=

lim

xa

f

(x)

?

lim

xa

g(x)

5.

xlima(

f

(x)g(x))

=

lim

xa

f

(x)

?

lim

xa

g(x)

6.

lim

xa

f (x) g(x)

=

lim

xa

f

(x)

lim

xa

g(x)

provided

lim

xa

g(x)

=

0

The Theorem also hold for one-sided limits and, with a little care,1 for infinite limits. For example,

if lim f (x) = -3 and lim g(x) = -, then

x2

x2

lim[ f (x) + g(x)] = -

x2

and lim f (x)g(x) =

x2

Corollary.

Suppose

that

p(x)

=

cn xn

+ cn-1xn-1

+ ? ? ? + c1x

+ c0

is

a

polynomial, then

lim

xa

p(x)

=

p(a).

Moreover, if r is any rational function, and a dom(r), then xlima r(x) = f (a).

Proof. Simply calculate:

lim

xa

p(x)

=

lim

xa

cn xn

+

?

?

?

+

c1

lim

xa

x

+

lim

xa

c0

=

cn

lim

xa

xn

+

?

?

?

+

c1

lim

xa

x

+

c0

= cnan + ? ? ? + c1a + c0

= p(a)

(rule 4) (rule 3) (rule 5 (repeatedly) and 2)

If

r

is

rational,

then

r(x)

=

p(x) q(x)

for

some

polynomials

p, q.

Rule

6

now

finished

things

off.

1If you end up with an indeterminate later using l'Ho^ pital's Rule.

form

0 0

,

,

-

,

etc.,

then

the

rules

don't

apply.

We

will

deal

with

these

limits

Examples

1.

Suppose

that

lim

xa

f

(x)

=

3,

lim

xa

g(x)

=

-1,

lim

xa-

h(x)

=

,

and

lim

xa+

h(x)

=

6.

Then

lim

xa

f

(x)

+

3g(x)

=

3

+

3(-1)

=

0

lim 2 f (x)g(x)h(x) = 2 ? 3 ? (-1) lim h(x) = -

xa-

xa-

lim 2 f (x)g(x)h(x) = 2 ? 3 ? (-1) ? 6 = 36

xa+

lim

xa-

g(x) f (x)h(x)

=

3

-1 lim h(x)

=

0

xa-

lim

xa+

g(x) f (x)h(x)

=

-1 3?6

=

-

1 18

2.

Simple evaluation:

lim

x1

x3

+ 2x2 - x 4x2 - 1

-

1

=

1+2-1-1 4-1

=

1 3

3.

Factorizing:

lim

x2-

x2 - 7x + 10 x2 - 4x + 4

=

lim

x2-

(x (x

- -

2)(x 2)(x

- -

5) 2)

= lim

x2-

x-5 x-2

= -3 lim

x2-

1 x-2

=

Roots and Rationalizing

Theorem.

lim

xa

f (x)

=

L

=

lim

xa

n

f (x) = n L

Recall how you would convert an expression with surds in the denominator into one with surds

in the numerator:

3

+45

=

3

+45

?

3 3

- -

55

=

4(3 - 5) 9-5

=

3

-

5

A similar approach can be used for limits.

Examples

1.

lim

x0

x

+

3 x

-

3

yields the indeterminate form

0 0

.

Multiplying by 1 =

xx

+ +

3 3

+ +

33

= 1 fixes

the problem:

lim

x0

x

+3 x

-

3

=

lim

x0

x

+3- x

3

?

xx

+ +

3 3

+ +

33

=

lim

x0

x(xx++33-+33)

=

lim

x0

x

1 +3+

3

=

21 3

2.

lim

x4

x2 + 9 - x-4

5

=

4 5

Comparing Limits and the Squeeze Theorem

While simple limits can be computed using the basic limit laws, more complicated functions are often best treated by comparison.

Theorem. Suppose that f (x) g(x) for all x = a and suppose that xlima f (x) and xlima g(x) both exist. Then

lim

xa

f

(x)

lim

xa

g(x)

Theorem

(Squeeze

Theorem).

Suppose

that

f (x)

g(x)

h(x)

for

all

x

=

a,

and

that

lim

xa

f

(x)

=

xlima h(x) = L. Then xlima g(x) exists and also equals L.

y

h(x)

g(x)

f (x)

a

x

y

Example In this example we compare the complicated

1

function f (x) = | that

xg|(. xS)in=cex-s1inx12sinwix1t2h

the much simpler 1 for all x = 0,

function we have

- |x| x sin

1 x2

|x|

Since lim |x| = 0 it follows that

x0

-1

1

x

lim x sin

x0

1 x2

=0

-1

Homework

1.

(a) (b)

PHreonvceetohratotxh-erwy =ise(cxo1/m3p-utye1/t3h)e(xli2m/3it+xlimx18/33xyx1-/-38+2 y2/3).

2. Suppose that f (x) < g(x) for all x = a and that limits of f and g both exist at x = a. Give

an

example

which

shows

that

we

may

only

conclude

that

lim

xa

f (x)

lim

xa

g(x).

That is, the

inequality need not be strict.

3.

Show

that

lim

x0

x2

cos(

1 x

)

exists

and

compute

it.

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