Question

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

Answer 1

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

Let $x=\frac{\pi }{4}+y$

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

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

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

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

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

$[\because tan\frac{\pi }{4}=1]$

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

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

$=-2\times 1\frac{1}{(1-0)}=-2$