Question

${lim}_{x\to \frac{\pi }{4}\frac{\sqrt{cosx}-\sqrt{sinx}}{x-\frac{\pi }{4}}}$

Answer 1

${lim}_{x\to \frac{\pi }{4}\frac{\sqrt{cosx}-\sqrt{sinx}}{x-\frac{\pi }{4}}}$

$={lim}_{x-\frac{\pi }{4}\to 0}\frac{(\sqrt{cosx}-\sqrt{sinx})}{(x-\frac{\pi }{4})}\times \frac{(\sqrt{cosx}+\sqrt{sinx})}{(\sqrt{cosx}+\sqrt{sinx})}$

$={lim}_{x-\frac{\pi }{4}\to 0}\frac{(cosx-sinx)}{(x-\frac{\pi }{4})(\sqrt{cosx}+\sqrt{sinx})}$

As $x\to \frac{\pi }{4}\Rightarrow x-\frac{\pi }{x}\to 0\Rightarrow$ let $x-\frac{\pi }{4}=y\Rightarrow y\to 0$

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

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

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

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

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

$=-\sqrt{2}\times \frac{1}{{\left(\frac{1}{\sqrt{2}}\right)}^{\frac{1}{2}}+{\left(\frac{1}{\sqrt{2}}\right)}^{\frac{1}{2}}}$ $[\because cos\frac{\pi }{4}=sin\frac{\pi }{4}=\frac{1}{\sqrt{2}}]$

$=\frac{-\sqrt{2}}{{\left(\frac{1}{\sqrt{2}}\right)}^{\frac{1}{2}}(1+1{)}^{\frac{1}{2}}}=\frac{-\sqrt{2}}{\sqrt{2}{\left(\frac{1}{\sqrt{2}}\right)}^{\frac{1}{2}}=\frac{-1}{2^{\frac{1}{4}}}}$