Question

evaluate the limit $lim_{x
ightarrow0}\frac{sin(2x)}{8x}=$

Explanation:

Step1: Use the limit formula $\lim_{u

ightarrow0}\frac{\sin u}{u}=1$
Let $u = 2x$. As $x
ightarrow0$, then $u
ightarrow0$. And $\frac{\sin(2x)}{8x}=\frac{1}{4}\cdot\frac{\sin(2x)}{2x}$.

Step2: Evaluate the limit

$\lim_{x
ightarrow0}\frac{\sin(2x)}{8x}=\frac{1}{4}\lim_{x
ightarrow0}\frac{\sin(2x)}{2x}$. Since $\lim_{u
ightarrow0}\frac{\sin u}{u} = 1$ and here $u = 2x$, $\lim_{x
ightarrow0}\frac{\sin(2x)}{2x}=1$. So $\frac{1}{4}\lim_{x
ightarrow0}\frac{\sin(2x)}{2x}=\frac{1}{4}\times1$.

Answer:

$\frac{1}{4}$