Question

The value of $\displaystyle \lim _{x \rightarrow 0} \frac{\cos h x-\cos x}{x \sin x}$ is

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

Answer 1

Answer: (A)

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