Question

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

Answer 1

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

Let $x=\frac{\pi }{4}+h\Rightarrow h=x-\frac{\pi }{4}$

as $x-\frac{\pi }{4},h\to 0$

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

$={lim}_{h\to 0}\frac{\sqrt{2}-(cos\frac{\pi }{4}cosh-sin\frac{\pi }{4}sinh)-(sin\frac{\pi }{4}cosh+cos\frac{\pi }{4}sinh)}{h^2}$

$={lim}_{h\to 0}\frac{\sqrt{2}-\left(\frac{cosh}{\sqrt{2}-\frac{sinh}{\sqrt{2}}}\right)-(\frac{cosh}{\sqrt{2}}+\frac{sinh}{\sqrt{2}})}{h^2}$

$={lim}_{h\to 0}\frac{\sqrt{2}-\frac{cosh}{\sqrt{2}}+\frac{sinh}{\sqrt{2}}-\frac{cosh}{\sqrt{2}}-\frac{sinh}{\sqrt{2}}}{h^2}={lim}_{h\to 0}\frac{\sqrt{2}-\sqrt{2}cosh}{h^2}$

$={lim}_{h\to 0}\frac{\sqrt{2}-(1-cosh)}{h^2}={lim}_{h\to 0}\frac{\sqrt{2}\left(2si{n}^2\frac{h}{2}\right)}{h^2}$

$={lim}_{h\to 0}2\sqrt{2}{\left(\frac{sin\frac{h}{2}}{\frac{h}{2}}\right)}^2\frac{1}{4}=2\sqrt{2}(1{)}^2\times \frac{1}{4}=\frac{1}{\sqrt{2}}$