Question

If the function $f$ defined on $\left(\frac{\pi}{6}, \frac{\pi}{3}\right)$ by
$f(x) = \begin{cases} \frac{\sqrt{2} cos\,x - 1}{cot\,x -1}, &x \ne \frac{\pi}{4} \\[2ex] k, &x = \frac{\pi}{4} \end{cases}$ is continuous, then $k$ is equal to

Answer 1

Since,$ f(x)$ is continuous, then
$\lim\limits_{x \to \frac{\pi}{4}} f(x) = f (\frac{\pi}{4})$
$\lim\limits_{x \to \frac{\pi}{4}} \frac{\sqrt{2} cos\,x - 1}{cot\,x - 1} = k$
Now by L- Hospital’s rule
$\lim\limits_{x \to \frac{\pi}{4}} \frac{\sqrt{2} \,sin \,x}{cosec^2 \,x} = k$
$\Rightarrow \frac{\sqrt{2}(\frac{1}{\sqrt{2}})}{(\sqrt{2})^2} = k$
$\Rightarrow k = \frac{1}{2}$