Question

If $f\left(x\right)=\frac{\tan (\frac{\pi }{4}-x)}{\cos 2x}$ for $x\ne \frac{\pi }{4}$ , then find the value which can be given to $f\left(x\right)$ at $x=\frac{\pi }{4}$ so that the function becomes continuous every where in $(0,\pi /2)$

Answer 1

For$f$ to be continuous at given point we must have,

$f\left(\frac{\pi }{4}\right)=\underset{x\to \frac{\pi }{4}}{lim}\frac{\tan (\frac{\pi }{4}-x)}{\cos 2x}$
Clearly form of limit is $\frac{0}{0}$. Thus applying L Hospital's rule.
$=\underset{x\to \frac{\pi }{4}}{lim}\frac{{\sec }^2(\frac{\pi }{4}-x)(-1)}{-2\sin 2x}=\frac{1}{2}$