Question

${lim}_{x\to \frac{\pi }{4}}\frac{\sqrt{2}-cosx-sinx}{(4x-\pi {)}^2}$

Answer 1

${lim}_{x\to \frac{\pi }{4}}\frac{\sqrt{2}-cosx-sinx}{(4x-\pi {)}^2}$

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

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

$={lim}_{y\to 0}\frac{\sqrt{2}-(cosycos\frac{\pi }{4}-sinysin\frac{\pi }{4})-(sinycos\frac{\pi }{4}+cossin\frac{\pi }{4})}{y^2}$

$=\frac{1}{16}{lim}_{y\to 0}\frac{\sqrt{2}(cosy\times \frac{1}{\sqrt{2}-siny+\frac{1}{\sqrt{2}}})-(\frac{siny}{\sqrt{2}}+\frac{cosy}{\sqrt{2}})}{y^2}$

$=\frac{1}{16}{lim}_{y\to 0}\frac{\sqrt{2}-\frac{1}{\sqrt{2}}(cosy-siny)-\frac{1}{\sqrt{2}(siny+cosy)}}{y^2}$

$=\frac{1}{16}{lim}_{y\to 0}\frac{\sqrt{2}-\frac{1}{\sqrt{2}\left[\right(cosy-siny)-(siny+cosy\left)\right]}}{y^2}$

$=\frac{1}{16}{lim}_{y\to 0}\frac{(\sqrt{2}-\frac{1}{\sqrt{2}cosy}+\frac{1}{\sqrt{2}siny}-\frac{1}{\sqrt{2}siny}-\frac{1}{\sqrt{2}cosy})}{y^2}$

$=\frac{1}{16}{lim}_{y\to 0}\frac{\sqrt{2}+\frac{2}{\sqrt{2}}cosy}{y^2}==\frac{1}{16}{lim}_{y\to 0}\frac{\sqrt{2}(1-cosy)}{y^2}$

$=\frac{\sqrt{2}}{16}{lim}_{y\to 0}\frac{2si{n}^2\frac{y}{2}}{y^2}=\frac{\sqrt{2}}{8}{lim}_{y\to 0}{\left(\frac{sin\frac{y}{2}}{\frac{y}{2}}\right)}^2\times \frac{1}{4}=\frac{1}{16\sqrt{2}}$