Question

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

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

Answer 1

Answer: (C)

$1$

Let $l=\underset{x\to 0}{lim}\frac{\cos hx-\cos x}{x\sin x}$ $\left[\frac{0}{0}form\right]$

Now, applying L'Hospital's rule, we get

$l=\underset{x\to 0}{lim}\frac{\sin hx+\sin x}{x\cos x+\sin x}$ $\left[\frac{0}{0}form\right]$

$[\because \frac{d}{dx}\left(\cos hx\right)=\sin hx]$

Again, applying L'Hospital's rule, we get

$l=\underset{x\to 0}{lim}\frac{\cos hx+\cos x}{x(-\sin x)+\cos x+\cos x}$

$[\because \frac{d}{dx}\left(\sin hx\right)=\cos hx]$

$\Rightarrow l=\underset{x\to 0}{lim}\frac{\cos hx+\cos x}{2\cos x-x\sin x}=\frac{\cos h\left(0\right)+\cos \left(0\right)}{2\cos \left(0\right)-0}$

$=\frac{1+1}{2}=1$ $[\because \cos h\left(0\right)=1and\cos \left(0\right)=1]$

$\Rightarrow l=1$

Alternate Method

Consider,

$\underset{x\to 0}{lim}\frac{\cos hx-\cos x}{x\sin x}=\underset{x\to 0}{lim}\left[\frac{\frac{e^x+e^{-x}}{2}-\cos x}{x\sin x}\right]$

$=\underset{x\to 0}{lim}\left[\frac{(1+\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots )-(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots )}{x(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots )}\right]$

$[\because e^x+e^{-x}=2(1+\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots )\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots and\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots ]$

$=\underset{x\to 0}{lim}\left[\frac{2(\frac{x^2}{2!}+\frac{x^6}{6!}+\cdots )}{x^2(1-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+\cdots )}\right]$

$=\underset{x\to 0}{lim}\left[\frac{2(\frac{1}{2!}+\frac{x^4}{6!}+\cdots )}{(1-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+\cdots )}\right]$

$=\frac{2(\frac{1}{2!}+0+0+\cdots )}{(1-0+0+\cdots )}=2\times \frac{1}{2}=1$.