Question

The value of $\underset{x\to 0}{lim}\frac{\sin mx-\sin nx}{x-\sin nx}$ is

• A. $\frac{m}{n}$
• B. $0$
• C. $-\frac{m-1}{1-n}$
• D. $\frac{m-n}{1-n}$

Answer 1

The correct option is D $\frac{m-n}{1-n}$

$\underset{x\to 0}{lim}\frac{x-\sin nx}{x-\sin nx}-\frac{x-\sin mx}{x-\sin nx}\to \frac{0}{0}\text{form}$
Apply L'hospital
$\underset{x\to 0}{lim}1-\frac{1-m\cos mx}{1-n\cos nx}$
$\underset{x\to 0}{lim}1-\frac{1-m}{1-n}$
$\underset{x\to 0}{lim}\frac{1-n-1+m}{1-n}$
$=\frac{m-n}{1-n}$