Question

$f\left(x\right)=\{\begin{array}{cc}\frac{\sin x-\cos x}{x-\frac{pi}{4}}& x\ne \frac{\pi }{4}\\ k& x=\frac{\pi }{4}\end{array}$.
If $f$ is continuous at $x=\frac{\pi }{4}$, find $k$.

Answer 1

Given $f\left(x\right)=\{\begin{array}{cc}\frac{\sin x-\cos x}{x-\pi /4},& x\ne \pi /4\\ k,& x=\pi /4\end{array}$ is continuous at $x=\pi /4$

$\Rightarrow \underset{x\to \pi /4^{-}}{lim}f\left(x\right)=\underset{x\to \pi /4^{+}}{lim}f\left(x\right)=f\left(\frac{\pi }{4}\right)=k$ ....(1)

$\underset{x\to \pi /4^{-}}{lim}f\left(x\right)=\underset{h\to 0}{lim}f(\frac{\pi }{4}-h)$

$\underset{h\to 0}{lim}\frac{\sin (\pi /4-h)-\cos (\pi /4-h)}{\frac{\pi }{4}-h-\frac{\pi }{4}}$

$=\underset{h\to 0}{lim}\frac{\sin (\pi /4-h)-\cos (\pi /4-h)}{-h}$

$=\underset{h\to 0}{lim}\frac{\sin \left(\pi /4\right)\cos h-\cos \pi /4\sin h-\cos \pi /4\cos h-\sin \pi /4\sin h}{-h}$

$\left(\begin{array}{c}\therefore \sin (A-B)=\sin A\cos B-\cos A\sin B\\ \cos (A-B)=\cos A\cos B+\sin A\sin B\end{array}\right)$

$\underset{h\to 0}{lim}\frac{\cos h/\sqrt{2}-\sin h/\sqrt{2}-\cos h/\sqrt{2}-\sin h/\sqrt{2}}{-h}$

$=\underset{h\to 0}{lim}\frac{-2\sin h}{\sqrt{2}-h}$

$=\sqrt{2}\underset{h\to 0}{lim}\frac{\sin h}{h}$

$=\sqrt{2}.1$

$=\sqrt{2}$

$\therefore \underset{x\to \pi /4^{-}}{lim}f\left(x\right)=\sqrt{2}$ .....(2)

(1) & (2) $\Rightarrow k=\sqrt{2}$

$\therefore k=\sqrt{2}$