Question

use the relation $lim_{\theta
ightarrow0}\frac{sin\theta}{\theta}=1$ to determine the limit. $lim_{\theta
ightarrow0}\frac{9sinsqrt{6}\theta}{sqrt{6}\theta}$

Explanation:

Step1: Factor out the constant

We know that for a limit $\lim_{\theta
ightarrow a}k\cdot f(\theta)=k\cdot\lim_{\theta
ightarrow a}f(\theta)$ where $k$ is a constant. Here $k = 9$, so $\lim_{\theta
ightarrow0}\frac{9\sin\sqrt{6}\theta}{\sqrt{6}\theta}=9\lim_{\theta
ightarrow0}\frac{\sin\sqrt{6}\theta}{\sqrt{6}\theta}$.

Step2: Apply the given - limit formula

Let $x=\sqrt{6}\theta$. As $\theta
ightarrow0$, then $x
ightarrow0$. We are given that $\lim_{x
ightarrow0}\frac{\sin x}{x}=1$. So $9\lim_{\theta
ightarrow0}\frac{\sin\sqrt{6}\theta}{\sqrt{6}\theta}=9\times1$.

Answer:

$9$