Question

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

Answer 1

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

Let $x=\frac{\pi }{2}+y$

$\Rightarrow y=x-\frac{\pi }{2}$

as $x\to \frac{\pi }{2},y\to 0$

$={lim}_{y\to 0}\frac{1-sin(\frac{\pi }{2}-y)}{y^2}$

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

$=2{\left({lim}_{y\to 0}\frac{sin\frac{y}{2}}{\frac{y}{2}}\right)}^2\times \frac{1}{4}[\because {lim}_{y\to 0}\frac{sinx}{x}=1]$

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