Question

lim(x->0) (sin(8x))/(2x)

Explanation:

Step1: Use the limit - formula $\lim_{u

ightarrow0}\frac{\sin u}{u}=1$
Let $u = 8x$. As $x
ightarrow0$, then $u
ightarrow0$. We can rewrite $\lim_{x
ightarrow0}\frac{\sin8x}{2x}$ as $\lim_{x
ightarrow0}\frac{\sin8x}{8x}\times\frac{8x}{2x}$.

Step2: Simplify the expression

$\lim_{x
ightarrow0}\frac{\sin8x}{8x}\times\frac{8x}{2x}=\lim_{x
ightarrow0}\frac{\sin8x}{8x}\times4$.
Since $\lim_{u
ightarrow0}\frac{\sin u}{u} = 1$ (here $u = 8x$), and $\lim_{x
ightarrow0}\frac{\sin8x}{8x}=1$.
So $\lim_{x
ightarrow0}\frac{\sin8x}{8x}\times4=1\times4$.

Answer:

$4$