Question

Find the value of the limit

$\underset{x\to \pi }{lim}\frac{\sqrt{2+\cos x}-1}{{\left(\pi -x\right)}^2}$ =

Answer 1

${lim}_{x\to \pi }\frac{\sqrt{2+\cos x}-1}{(\pi -x{)}^2}\left(\frac{0}{0}\right)$ form

Apply L-hospital rule

${lim}_{x\to \pi }\frac{\sqrt{2+\cos x-1}}{(\pi -x{)}^2}\left(\frac{0}{0}\right)$ form

$={lim}_{x\to \pi }\frac{1(-\sin x)}{2\sqrt{2+\cos x}\left(2\right(\pi -x\left)\right)(-1)}$

$={lim}_{x\to \pi }\frac{\sin x}{4(\pi -x)\sqrt{2+\cos x}}$

$\begin{array}{cc}\Rightarrow & \underset{x\to \pi }{lim}\frac{\cos x}{-4\sqrt{2+\cos x+4(\pi -x)}}\frac{1(-\sin x)}{2\sqrt{2+\cos x}}\\ & =\frac{-1}{-4}=\frac{1}{4}\end{array}$