Question

use $lim_{x
ightarrow0}\frac{sin x}{x}=1$ and/or $lim_{x
ightarrow0}\frac{cos x - 1}{x}=0$ to evaluate the following limit.
$lim_{x
ightarrow0}\frac{\tan4x}{sin9x}$
select the correct choice and, if necessary, fill in the answer box to complete your choice.
a. $lim_{x
ightarrow0}\frac{\tan4x}{sin9x}=square$ (type an integer or a simplified fraction.)
b. the limit is undefined

Explanation:

Step1: Rewrite tan 4x

Recall that $\tan 4x=\frac{\sin 4x}{\cos 4x}$. So, $\lim_{x
ightarrow0}\frac{\tan 4x}{\sin 9x}=\lim_{x
ightarrow0}\frac{\sin 4x}{\cos 4x\sin 9x}$.

Step2: Use the limit property $\lim_{u

ightarrow0}\frac{\sin u}{u} = 1$
We can rewrite the limit as $\lim_{x
ightarrow0}\frac{\sin 4x}{4x}\cdot\frac{4x}{\cos 4x\sin 9x}$. Then $\lim_{x
ightarrow0}\frac{\sin 4x}{4x}\cdot\frac{4x}{\cos 4x\sin 9x}=\lim_{x
ightarrow0}\frac{\sin 4x}{4x}\cdot\frac{4}{\cos 4x}\cdot\frac{x}{\sin 9x}$.
Since $\lim_{x
ightarrow0}\frac{\sin 4x}{4x} = 1$ and $\lim_{x
ightarrow0}\frac{x}{\sin 9x}=\lim_{x
ightarrow0}\frac{1}{9}\cdot\frac{9x}{\sin 9x}=\frac{1}{9}$ (because $\lim_{u
ightarrow0}\frac{\sin u}{u} = 1$ with $u = 9x$).

Step3: Calculate the limit

$\lim_{x
ightarrow0}\frac{\sin 4x}{4x}\cdot\frac{4}{\cos 4x}\cdot\frac{x}{\sin 9x}=1\cdot\frac{4}{\cos(0)}\cdot\frac{1}{9}$. Since $\cos(0)=1$, we have $1\cdot\frac{4}{1}\cdot\frac{1}{9}=\frac{4}{9}$.

Answer:

A. $\frac{4}{9}$