Question

${lim}_{x\to \frac{\pi }{4}}\frac{1-tanx}{1-\sqrt{2}sinx}$

Answer 1

${lim}_{x\to \frac{\pi }{4}}\frac{1-tanx}{1-\sqrt{2}sinx}$

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

$={lim}_{y\to 0}\frac{1-tan(y+\frac{\pi }{4})}{1-\sqrt{2}sin(y+\frac{\pi }{4})}$

$={lim}_{y\to 0}\frac{1-\left(\frac{tan\frac{\pi }{4}+tany}{1+tan\frac{\pi }{4}+tany}\right)}{1-\sqrt{2}(sinycos\frac{\pi }{4}+cosysin\frac{\pi }{4})}$

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

$=-2\frac{{lim}_{y\to 0}tany}{{lim}_{y\to 0}(1-tany)\times {lim}_{y\to 0}(1-siny-cosy)}$

$=\frac{-2\left({lim}_{y\to 0}\frac{tany}{y}\right)\times y}{\left({lim}_{y\to 0}\right(1)-{lim}_{y\to 0}tany)\times (1-{lim}_{y\to 0}\frac{siny}{y}\times y-cos0)}$

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

$={lim}_{y\to 0}\frac{2}{1-y}=2$