Question

1. $limlimits _{x\to 0}\frac {sin mx}{sin nx}$

Explanation:

Step1: Use the limit formula $\lim_{x

ightarrow0}\frac{\sin ax}{ax} = 1$

$$\lim_{x ightarrow0}\frac{\sin mx}{\sin nx}=\lim_{x ightarrow0}\frac{\sin mx}{mx}\cdot\frac{nx}{\sin nx}\cdot\frac{m}{n}$$

Step2: Apply the limit formula

Since $\lim_{x
ightarrow0}\frac{\sin mx}{mx} = 1$ and $\lim_{x
ightarrow0}\frac{nx}{\sin nx}=1$, we have:

$$\lim_{x ightarrow0}\frac{\sin mx}{mx}\cdot\frac{nx}{\sin nx}\cdot\frac{m}{n}=1\times1\times\frac{m}{n}$$

Answer:

$\frac{m}{n}$