Question

Let $f\left(x\right)=\frac{\tan (\frac{\pi }{4}-x)}{\cot 2x}$ for $x\ne \frac{\pi }{4}$ , then for f to be continuous at $x=\frac{\pi }{4},f\left(\frac{\pi }{4}\right)$ must be equal .

Answer 1

Let $f\left(n\right)=\frac{tan(\frac{\pi }{4}-x)}{cot2x}$ for $x\ne \pi /4$

$kx=\pi /4.$

function is continuous so

by L' Hospital.

$\underset{x\to \frac{\pi }{4}}{lim}\frac{(\frac{\pi }{4}-x)}{cos2x}=k$

$\underset{x\to \frac{\pi }{4}}{lim}\frac{-se{c}^2(\frac{\pi }{4}-x)}{-2sin2x}=k$

$=\underset{h\to 0}{lim}+\frac{se{c}^2(\frac{\pi }{2}-\frac{\pi }{2}+h)}{2sin(\frac{\pi }{4}-2h)}=k$

$\Rightarrow \underset{h\to 0}{lim}+\frac{se{c}^{2}h}{2cos2h}=k$

$\Rightarrow k=\frac{1}{2}$