Question

find the limit using $lim_{\theta
ightarrow0}\frac{sin\theta}{\theta}=1$.
$lim_{x
ightarrow0}\frac{xcsc9x}{cos11x}$
select the correct choice below and, if necessary, fill in the answe
a. $lim_{x
ightarrow0}\frac{xcsc9x}{cos11x}=square$ (simplify your answer.)
b. the limit does not exist

Explanation:

Step1: Rewrite csc function

Recall that $\csc 9x=\frac{1}{\sin 9x}$, so the limit $\lim_{x
ightarrow0}\frac{x\csc 9x}{\cos 11x}=\lim_{x
ightarrow0}\frac{x}{\sin 9x\cos 11x}$.

Step2: Rewrite the limit

We can rewrite $\lim_{x
ightarrow0}\frac{x}{\sin 9x\cos 11x}$ as $\lim_{x
ightarrow0}\frac{x}{\sin 9x}\cdot\frac{1}{\cos 11x}$.

Step3: Use the limit formula

Let $u = 9x$, then as $x
ightarrow0$, $u
ightarrow0$. And $\lim_{x
ightarrow0}\frac{x}{\sin 9x}=\frac{1}{9}\lim_{x
ightarrow0}\frac{9x}{\sin 9x}$. Since $\lim_{\theta
ightarrow0}\frac{\sin\theta}{\theta}=1$, then $\lim_{\theta
ightarrow0}\frac{\theta}{\sin\theta}=1$. So $\frac{1}{9}\lim_{x
ightarrow0}\frac{9x}{\sin 9x}=\frac{1}{9}$. Also, $\lim_{x
ightarrow0}\frac{1}{\cos 11x}=\frac{1}{\cos(0)} = 1$.

Step4: Calculate the limit

$\lim_{x
ightarrow0}\frac{x}{\sin 9x}\cdot\frac{1}{\cos 11x}=\frac{1}{9}\times1=\frac{1}{9}$.

Answer:

A. $\frac{1}{9}$