Question

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

• A. $1$
• B. $\frac{1}{2}$
• C. $\frac{1}{3}$
• D. $2$

Answer 1

The correct option is A $1$

$\underset{x\to 0}{lim}\frac{\cos hx-\cos x}{x^2}\cdot \frac{x}{\sin x}$
$=\underset{x\to 0}{lim}\frac{\cos hx-\cos x}{x^2}\cdot 1$

The above limit is of the form $\frac{0}{0}$
Using L'Hospitals' Rule,

$\underset{x\to 0}{lim}\frac{\cos hx-\cos x}{x^2}=\underset{x\to 0}{lim}\frac{\sin hx+\sin x}{2x}$
$=\underset{x\to 0}{lim}\frac{\cos hx+\cos x}{2}=\frac{1+1}{2}=1$