Question

1. 0 / 1.33 points 1/2 submissions used

find the limit.
\\(\lim_{x\to0}\frac{\sin(x)}{\sin(\pi x)}\\)

Explanation:

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

ightarrow0}\frac{\sin(u)}{u}=1$
Rewrite the given limit $\lim_{x
ightarrow0}\frac{\sin(x)}{\sin(\pi x)}$ as $\lim_{x
ightarrow0}\frac{\sin(x)}{x}\cdot\frac{\pi x}{\sin(\pi x)}\cdot\frac{1}{\pi}$.

Step2: Apply the limit - formula

We know that $\lim_{x
ightarrow0}\frac{\sin(x)}{x} = 1$ and $\lim_{x
ightarrow0}\frac{\pi x}{\sin(\pi x)}=1$. So, $\lim_{x
ightarrow0}\frac{\sin(x)}{x}\cdot\frac{\pi x}{\sin(\pi x)}\cdot\frac{1}{\pi}=1\times1\times\frac{1}{\pi}=\frac{1}{\pi}$.

Answer:

$\frac{1}{\pi}$