Question

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

Answer 1

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

$\Rightarrow x\to \pi ,$ then $x-\pi \to 0$ or let $x-\pi =y\Rightarrow y\to 0$

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

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

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

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

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

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

$[\because {lim}_{0\to 0}\frac{sin\theta }{\theta }=1]$

$=\frac{1}{4}$