Question

${lim}_{x\to \frac{\pi }{2}}\frac{(\frac{\pi }{2}-x)sinx-2cosx}{(\frac{\pi }{2}-x)+cotx}$

Answer 1

${lim}_{x\to \frac{\pi }{2}}\frac{(\frac{\pi }{2}-x)sinx-2cosx}{(\frac{\pi }{2}-x)+cotx}$

Let $x=\frac{\pi }{2}+y\Rightarrow y=x-\frac{\pi }{2}$ as $x\to \frac{\pi }{2},y\to 0$

$={lim}_{y\to 0}\frac{(ysin(\frac{\pi }{2}-y)-2cos(\frac{\pi }{2}-y))}{y+cot(\frac{\pi }{2}-y)}={lim}_{y\to 0}\left(\frac{ycosy-2siny}{1+tany}\right)$

$={lim}_{y\to 0}\left(\frac{cosy-2\frac{siny}{y}}{1+\frac{tany}{y}}\right)=\frac{{lim}_{y\to 0}cosy-2{lim}_{y\to 0}\frac{siny}{y}}{1+{lim}_{y\to 0}\frac{tany}{y}}$ $[\because {lim}_{y\to 0}\frac{siny}{y}=1,{lim}_{y\to 0}\frac{tany}{y}=1]$

$=\frac{1-2}{1+1}=\frac{-1}{2}=-\frac{1}{2}$