evaluate the limit $lim_{x\rightarrow0}\frac{sin(2x)}{8x}=$

Explanation:

Step1: Use the limit formula $\lim_{u\rightarrow0}\frac{\sin u}{u}=1$

Let $u = 2x$. As $x\rightarrow0$, then $u\rightarrow0$. And $\frac{\sin(2x)}{8x}=\frac{1}{4}\cdot\frac{\sin(2x)}{2x}$.

Step2: Evaluate the limit

$\lim_{x\rightarrow0}\frac{\sin(2x)}{8x}=\frac{1}{4}\lim_{x\rightarrow0}\frac{\sin(2x)}{2x}$. Since $\lim_{u\rightarrow0}\frac{\sin u}{u} = 1$ and here $u = 2x$, $\lim_{x\rightarrow0}\frac{\sin(2x)}{2x}=1$. So $\frac{1}{4}\lim_{x\rightarrow0}\frac{\sin(2x)}{2x}=\frac{1}{4}\times1$.

Answer:

$\frac{1}{4}$