Question

evaluate:
$lim_{\theta
ightarrow0}2\frac{sin(9\theta)}{\tan(8\theta)}=square$
give your answer as a rational number.
enter here (using math notation or by attaching in an image) an explanation of your solu

Explanation:

Step1: Rewrite tangent

Recall that $\tan(x)=\frac{\sin(x)}{\cos(x)}$. So, $\lim_{\theta
ightarrow0}2\frac{\sin(9\theta)}{\tan(8\theta)}=\lim_{\theta
ightarrow0}2\frac{\sin(9\theta)\cos(8\theta)}{\sin(8\theta)}$.

Step2: Use limit - formula $\lim_{u

ightarrow0}\frac{\sin(u)}{u} = 1$
Let $u = 9\theta$ and $v=8\theta$. We can rewrite the limit as $2\lim_{\theta
ightarrow0}\frac{\sin(9\theta)}{9\theta}\times9\times\frac{8\theta}{\sin(8\theta)}\times\frac{\cos(8\theta)}{8}$.
As $\theta
ightarrow0$, $\lim_{\theta
ightarrow0}\frac{\sin(9\theta)}{9\theta}=1$ and $\lim_{\theta
ightarrow0}\frac{8\theta}{\sin(8\theta)} = 1$, and $\lim_{\theta
ightarrow0}\cos(8\theta)=\cos(0) = 1$.

Step3: Calculate the limit value

$2\times1\times9\times1\times\frac{1}{8}=\frac{9}{4}$.

Answer:

$\frac{9}{4}$