Question

If $f\left(x\right)=\frac{\sqrt{2}\cos x-1}{\cot x-1},x\ne \frac{\pi }{4}$ then, find the value of $f\left(\frac{\pi }{4}\right)$ so that $f\left(x\right)$ becomes continuous at $x=\frac{\pi }{4}$

Answer 1

$f\left(x\right)=\frac{\sqrt{2}\cos x-1}{\cot x-1},x\ne \frac{x}{4}$

$\underset{x\to \frac{\pi }{4}}{lim}f\left(x\right)\underset{x\to \frac{\pi }{4}}{lim}\frac{\sqrt{2}\cos x-1}{\cot x-1}$ $\frac{0}{0}$ form

$=\underset{x\to \frac{\pi }{4}}{lim}\frac{-\sqrt{2}\sin x}{-cose{c}^{2}x}$

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

$\underset{x\to \frac{\pi }{4}}{lim}=\frac{1}{2}$

so $f\left(x\right)$ to be continuous at $x=\frac{\pi }{4}$

$\underset{x\to \frac{\pi }{4}}{lim}f\left(x\right)$ has to be equal to $f\left(\frac{\pi }{4}\right)$

$\therefore f\left(\frac{\pi }{4}\right)=\frac{1}{2}$