5 Steps to a 5 AP Calculus BC 2019

Limits and Continuity 55

Squeeze Theorem

If f, g, andh are functions defined on some open interval containinga such that
g(x)≤ f(x)≤h(x) for allxin the interval except possibly ataitself, and limx→ag(x)=

limx→ah(x)=L, thenlimx→a f(x)=L.

Theorems on Limits

(1) limx→ 0
sinx
x
=1 and (2) limx→ 0
cosx− 1
x

= 0

Example 1

Find the limit if it exists: limx→ 0
sin 3x
x

.

Substituting 0 into the expression would lead to 0/0. Rewrite
sin 3x
x
as

3

3

·

sin 3x

x

and

thus, limx→ 0
sin 3x
x
=xlim→ 0
3 sin 3x
3 x
=3limx→ 0
sin 3x
3 x
.Asxapproaches 0, so does 3x. Therefore,

3limx→ 0
sin 3x
3 x
=3 lim 3 x→ 0
sin 3x
3 x
=3(1)=3. (Note that lim 3 x→ 0
sin 3x
3 x
is equivalent to limx→ 0
sinx
x
by

replacing 3xbyx.) Verify your result with a calculator. (See Figure 5.1-7.)

[–10,10] by [–4,4]

Figure 5.1-7

Example 2

Find the limit if it exists: limh→ 0
sin 3h
sin 2h

.

Rewrite
sin 3h
sin 2h

as

3

(

sin 3h

3 h

)

2

(

sin 2h

2 h

).Ash approaches 0, so do 3h and 2h. Therefore,

limh→ 0
sin 3h
sin 2h

=

3 lim 3 h→ 0

sin 3h

3 h

2 lim 2 h→ 0

sin 2h

2 h

=

3(1)

2(1)

=

3

2

. (Note that substitutingh=0 into the original

expression would have produced 0/0). Verify your result with a calculator. (See Figure
5.1-8.)