Question

use the relation $lim_{\theta
ightarrow0}\frac{sin\theta}{\theta}=1$ to determine the limit.
$lim_{\theta
ightarrow0}\frac{sin\theta}{sin(3\theta)}$
select the correct answer below and, if necessary, fill in the answer box to complete your choice.
a. $lim_{\theta
ightarrow0}\frac{sin\theta}{sin(3\theta)}=square$ (type an integer or a simplified fraction.)
b. the limit does not exist.

Explanation:

Step1: Rewrite the limit

We know that $\sin(3\theta)=3\theta\cdot\frac{\sin(3\theta)}{3\theta}$. So, $\lim_{\theta
ightarrow0}\frac{\sin\theta}{\sin(3\theta)}=\lim_{\theta
ightarrow0}\frac{\sin\theta}{3\theta\cdot\frac{\sin(3\theta)}{3\theta}}$.

Step2: Rearrange the terms

$\lim_{\theta
ightarrow0}\frac{\sin\theta}{3\theta\cdot\frac{\sin(3\theta)}{3\theta}}=\frac{1}{3}\lim_{\theta
ightarrow0}\frac{\frac{\sin\theta}{\theta}}{\frac{\sin(3\theta)}{3\theta}}$.

Step3: Apply the limit formula

Since $\lim_{\theta
ightarrow0}\frac{\sin\theta}{\theta} = 1$ and $\lim_{\theta
ightarrow0}\frac{\sin(3\theta)}{3\theta}=1$, we have $\frac{1}{3}\cdot\frac{1}{1}=\frac{1}{3}$.

Answer:

A. $\lim_{\theta
ightarrow0}\frac{\sin\theta}{\sin(3\theta)}=\frac{1}{3}$