Calculus- Early Transcendentals, 2021a

100 Limits
This limit is just as hard as sinx/x, but closely related to it, so that we don’t have to do a similar calculation;
instead we can do a bit of tricky algebra.
cosx − 1
x
= cosx − 1 cosx + 1
x cosx + 1 = cos2 x − 1
x(cosx + 1) =
−sin2 x
x(cosx + 1) = −sinx sinx
x cosx + 1 .
To compute the desired limit it is sufficient to compute the limits of the two final fractions, as x goes to 0.
The first of these is the hard limit we’ve just done, namely 1. The second turns out to be simple, because
the denominator presents no problem:
Thus,
lim
x→0
sinx
cosx + 1 =
sin0
cos0 + 1 = 0 2 = 0.
cosx − 1
lim = 0.
x→0 x
Example 3.38: Limit of Other Trigonometric Functions
sin5xcosx
Compute the following limit lim .
x→0 x
Solution. We have
sin5xcosx
lim
x→0 x
5sin5xcosx
= lim
x→0 5x
( ) sin5x
= lim 5cosx
x→0 5x
= 5 · (1) · (1) = 5
sin5x
since cos(0)=1andlim = 1. ♣
x→0 5x
Let’s do a harder one now.
Example 3.39: Limit of Other Trig Functions
Compute the following limit: lim
x→0
tan 3 2x
x 2 sin7x .