Question

The value of $\underset{x\to \pi }{lim}\frac{\sqrt{2+\cos x}-1}{{(\pi -x)}^2}$ is

• A. $\frac{1}{4}$
• B. $2$
• C. $\frac{1}{2}$
• D. Zero

Answer 1

Answer: (A)

$\frac{1}{4}$

Solution:

We have, ${lim}_{x\to \pi }\frac{\sqrt{2+\cos x}-1}{(\pi -x{)}^2}$

Substitute $x=\pi ,$ we get $\frac{0}{0}$ form, so by L-Hospital's Rule we have

$={lim}_{x\to \pi }\frac{\frac{-\sin x}{2\sqrt{2+\cos x}}}{-2(\pi -x)}$

Putting $x=\pi$, it is again in $\frac{0}{0}$ form, so again diffrentiating, we get

$={lim}_{x\to \pi }\frac{\frac{-\cos x}{2\sqrt{2+\cos x}}-\frac{{\sin }^{2}x}{2}(2+\cos x{)}^{-3/2}}{2}$

Putting $x=\pi$, we get

$=\frac{\frac{1}{2}+0}{2}=\frac{1}{4}$ ... $[\cos \pi =-1,\sin \pi =0]$

Hence, A is the correct option.