Question

If $f\left(x\right)=\frac{1-tanx}{1-\sqrt{2}sinx}$, for $x\ne \frac{\pi }{4}$ is continuous at $x=\frac{\pi }{4}$, find $f\left(\frac{\pi }{4}\right)$

Answer 1

$f\left(x\right)=\frac{1-\tan x}{1-\sqrt{2}\sin x}$ For $x\ne \frac{4\pi }{4}$

it is continuous at $x=\frac{\pi }{4}$

$\therefore f\left(\frac{\pi }{4-}\right)=f\left(\frac{\pi }{4+}\right)$

$f\left(\frac{\pi }{4}\right)={lim}_{x\to \frac{\pi }{4}}\frac{1-\tan x}{1-\sqrt{2}\sin x}$

It is in $\frac{0}{0}$ form

So L' hospital rule is applicable

$\Rightarrow f\left(\frac{\pi }{4}\right)={lim}_{x\to \frac{\pi }{4}}\frac{1-\tan x}{1-\sqrt{2}\sin x}$

$={lim}_{x\to \frac{\pi }{4}}\frac{-{\sec }^{2}x}{-2\cos x}$ [ by differentiation numerator and denominator ]

$={lim}_{x\to \frac{\pi }{4}}\frac{{\sec }^{2}x}{2\cos x}$

$=\frac{{\sec }^2\frac{\pi }{4}}{2\cos \frac{\pi }{4}}$ [ By putting the limit value ]

$=\frac{2}{2.\frac{1}{\sqrt{2}}}$

$=\sqrt{2}.$

$\therefore f\left(\frac{\pi }{4}\right)=\sqrt{2}.$

Hence, the answer is $\sqrt{2}.$