Question

use the relation $lim_{\theta
ightarrow0}\frac{sin\theta}{\theta}=1$ to determine the limit.
$lim_{\theta
ightarrow0}\frac{9sinsqrt{5}\theta}{sqrt{5}\theta}$
select the correct answer below and, if necessary, fill in the answer box to complete your choice.
a. $lim_{\theta
ightarrow0}\frac{9sinsqrt{5}\theta}{sqrt{5}\theta}=square$ (type an integer or a simplified fraction.)
b. the limit does not exist.

Explanation:

Step1: Factor out the constant

We have the limit $\lim_{\theta
ightarrow0}\frac{9\sin\sqrt{5\theta}}{\sqrt{5\theta}}$. We can rewrite it as $9\times\lim_{\theta
ightarrow0}\frac{\sin\sqrt{5\theta}}{\sqrt{5\theta}}$.

Step2: Apply the limit formula

Given that $\lim_{x
ightarrow0}\frac{\sin x}{x} = 1$. Here, let $x = \sqrt{5\theta}$. As $\theta
ightarrow0$, $x=\sqrt{5\theta}
ightarrow0$. So, $\lim_{\theta
ightarrow0}\frac{\sin\sqrt{5\theta}}{\sqrt{5\theta}}=1$.

Step3: Calculate the final limit

Since $9\times\lim_{\theta
ightarrow0}\frac{\sin\sqrt{5\theta}}{\sqrt{5\theta}}$, and $\lim_{\theta
ightarrow0}\frac{\sin\sqrt{5\theta}}{\sqrt{5\theta}} = 1$, then $9\times1=9$.

Answer:

A. $\lim_{\theta
ightarrow0}\frac{9\sin\sqrt{5\theta}}{\sqrt{5\theta}}=9$