Question

question 19
find the limit: $lim_{\theta
ightarrow0}\frac{sin(8\theta)}{\theta+\tan(5\theta)}$.
0
$-\frac{8}{5}$
$\frac{4}{3}$
$\frac{3}{4}$
$\frac{8}{5}$

Explanation:

Step1: Use limit - related formulas

We know that $\lim_{x
ightarrow0}\frac{\sin x}{x} = 1$ and $\lim_{x
ightarrow0}\frac{\tan x}{x}=1$.
Rewrite $\lim_{\theta
ightarrow0}\frac{\sin(8\theta)}{\theta+\tan(5\theta)}$ as $\lim_{\theta
ightarrow0}\frac{\sin(8\theta)}{\theta+\frac{\sin(5\theta)}{\cos(5\theta)}}$.
Multiply both the numerator and denominator by $\frac{1}{\theta}$:
$\lim_{\theta
ightarrow0}\frac{\frac{\sin(8\theta)}{\theta}}{\frac{\theta}{\theta}+\frac{\tan(5\theta)}{\theta}}=\lim_{\theta
ightarrow0}\frac{\frac{\sin(8\theta)}{\theta}}{1 + \frac{\tan(5\theta)}{\theta}}$.

Step2: Apply limit formulas

Since $\lim_{\theta
ightarrow0}\frac{\sin(8\theta)}{\theta}=\lim_{\theta
ightarrow0}\frac{\sin(8\theta)}{8\theta}\times8 = 8$ (let $x = 8\theta$, as $\theta
ightarrow0,x
ightarrow0$) and $\lim_{\theta
ightarrow0}\frac{\tan(5\theta)}{\theta}=\lim_{\theta
ightarrow0}\frac{\tan(5\theta)}{5\theta}\times5=5$ (let $x = 5\theta$, as $\theta
ightarrow0,x
ightarrow0$).
Substitute these values into the limit: $\frac{8}{1 + 5}=\frac{8}{6}=\frac{4}{3}$.

Answer:

$\frac{4}{3}$