Question

${lim}_{x\to \frac{\pi }{4}}\frac{cosx-sinx}{(\frac{\pi }{4}-x)(cosx+sinx)}$

Answer 1

${lim}_{x\to \frac{\pi }{4}}\frac{cosx-sinx}{(\frac{\pi }{4}-x)(cosx+sinx)}$

$\Rightarrow x\to \frac{\pi }{4},$ then $x-\frac{\pi }{4}\to 0$ let $x-\frac{\pi }{4}=y$ $\therefore y\to 0$

$={lim}_{y\to 0}\frac{cos(\frac{\pi }{4}+y)-sin(\frac{\pi }{4}+y)}{-y(cos(\frac{\pi }{4}+y)+sin(\frac{\pi }{4}+y))}$

$={lim}_{y\to 0}\frac{[(cos\frac{\pi }{4}cosy-sin\frac{\pi }{4}siny)-(sin\frac{\pi }{4}cosy+cos\frac{\pi }{4}siny)]}{-y(cos(\frac{\pi }{4}+y)+sin(\frac{\pi }{4}+y))}$

$={lim}_{y\to 0}\frac{[\frac{cosy}{\sqrt{2}}-\frac{siny}{\sqrt{2}}-\frac{cosy}{\sqrt{2}}-\frac{siny}{\sqrt{2}}]}{-y(cos(\frac{\pi }{4}+y)+sin(\frac{\pi }{4}+y))}={lim}_{y\to 0}\frac{\frac{-2siny}{\sqrt{2}}}{-y(cos(\frac{\pi }{4}+y)+sin(\frac{\pi }{4}+y))}$

$=\sqrt{2}{lim}_{y\to 0}\left(\frac{siny}{y}\right)\times \frac{1}{{lim}_{y\to 0}[cos(\frac{\pi }{4}+y)+sin(\frac{\pi }{4}+y)]}$

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