Calculus- Early Transcendentals, 2021a

3.6. A Trigonometric Limit 101
Solution. Recall that the tan 3 (2x) means that tan(2x) isbeingraisedtothethirdpower.
lim
x→0
tan 3 (2x)
x 2 sin(7x)
= lim
x→0
(sin(2x)) 3
x 2 sin(7x)cos 3 (2x)
= lim
x→0
x 2 (7x)
(2x) 3 ( sin(2x)
(
sin(7x)
7x
= lim
x→0
8x 3 (1) 3
7x 3 (1)(1 3 )
8
= lim
x→0 7
2x
) 3
)
cos 3 (2x)
Rewrite in terms of sin and cos
Make sine terms look like: sinθ
θ
sinnx
Replace lim with 1. Also,cos(0)=1.
x→0 nx
Cancel x 3 ’s.
= 8 7 . ♣
Example 3.40: Applying the Squeeze Theorem
( ) 1√x
Compute the following limit: lim
x→0 x3 cos .
+
Solution. We use the Squeeze Theorem to evaluate this limit. We know that cosα satisfies −1 ≤ cosα ≤ 1
for any choice of α. Therefore we can write:
( ) 1√x
−1 ≤ cos ≤ 1
Since x → 0 + implies x > 0, multiplying by x 3 gives:
( ) 1√x
−x 3 ≤ x 3 cos ≤ x 3 .
( ( )) 1√x
lim ) ≤ lim x 3 cos ≤ lim
x→0 +(−x3 x→0 + x→0 x3 .
+
But using our rules we know that
lim )=0,
x→0 +(−x3
lim
x→0 x3 = 0
+
and the Squeeze Theorem says that the only way this can happen is if
( ) 1√x
lim
x→0 x3 cos = 0.
+
♣