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

Answer: (A)

Let $I=\underset{x\to 0}{\lim }\frac{\cos hx-\cos x}{x\sin x}[\frac{0}{0}$ form $]$
Now, applying L'Hospital's rule, we get
$I=\underset{x\to 0}{\lim }\frac{\sin hx+\sin x}{x\cos x+\sin x}\left[\frac{0}{0}\text{ form }\right]$
$[\because \frac{d}{dx}(\cos hx)=\sin hx]$
Again, applying L'Hospital's rule, we get
$I=\underset{x\to 0}{\lim }\frac{\cos hx+\cos x}{x(-\sin x)+\cos x+\cos x}$
$\left[\because \frac{d}{dx}(\sin hx)=\cos hx\right]$
$\Rightarrow I=\underset{x\to 0}{\lim }\frac{\cos hx+\cos x}{2\cos x-x\sin x}=\frac{\cos h(0)+\cos (0)}{2\cos (0)-0}$
$=\frac{1+1}{2}=1$
$[\because \cos h(0)=1\text{ and }\cos (0)=1]$
$\Rightarrow I=1$