Question

${lim}_{x\to \pi }\frac{\sqrt{2+cosx}-1}{(\pi -x{)}^2}$

Answer 1

${lim}_{x\to \pi }\frac{\sqrt{2+cosx}-1}{(\pi -x{)}^2}$

$={lim}_{x\to \pi }\frac{\sqrt{2+cosx}-1}{(\pi -x{)}^2}\times \frac{\sqrt{2+cosx}+1}{\sqrt{2+cosx}+1}$

$={lim}_{x\to \pi }\frac{(2+cosx)-1}{(\pi -x{)}^2(\sqrt{2+cosx}+1)}$

$={lim}_{x\to \pi }\frac{1+cosx}{(\pi -x{)}^2(\sqrt{2+cosx}+1)}$

Let $\pi -x=y,x\to \pi ,y\to 0$

$={lim}_{y\to 0}\frac{1+cos(\pi -y)}{\sqrt{2+cos(\pi -y)}+1}$

$={lim}_{y\to 0}\frac{2si{n}^2\frac{y}{2}}{\sqrt{2-cosy}+1}$

$=2{lim}_{y\to 0}{\left(\frac{\frac{siny}{2}}{\frac{y}{2}}\right)}^2\times \frac{1}{4}\frac{1}{\sqrt{2-cosy}+1}$

$=2\times {\left({lim}_{y\to 0}\frac{siny}{2}\right)}^2\times \frac{1}{4}\frac{1}{{lim}_{y\to 0}\sqrt{2-cosy}+1}$

$=2\times 1\times \frac{1}{4}\frac{1}{\sqrt{2-1}+1}$

$=2\times 1\times \frac{1}{4}\frac{1}{1+1}$

$=2\times 1\times \frac{1}{4}\times \frac{1}{2}=\frac{1}{4}$