Question

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

Answer 1

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

$\Rightarrow x\to \frac{\pi }{2}$, then $x-\frac{\pi }{2}\to 0,x-\frac{\pi }{2}=y,\Rightarrow y\to 0$

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

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

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

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

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