# $\lim _{x \rightarrow \frac{\pi}{4}} \frac{\sin x-\cos x}{x-\frac{\pi}{4}}$

Question.

$\lim _{x \rightarrow \frac{\pi}{4}} \frac{\sin x-\cos x}{x-\frac{\pi}{4}}$

solution:

Given $\lim _{x \rightarrow \frac{\pi}{4}} \frac{\sin x-\cos x}{x-\frac{\pi}{4}}$

We have

$\sin x-\cos x=\sqrt{2}\left(\frac{\sin x}{\sqrt{2}}-\frac{\cos x}{\sqrt{2}}\right)=\sqrt{2}\left(\sin x \cos \left(\frac{\pi}{4}\right)-\cos x \sin \left(\frac{\pi}{4}\right)\right)$

By using this formula in given equation we get

$\Rightarrow \sqrt{2}\left(\sin x \cos \left(\frac{\pi}{4}\right)-\cos x \sin \left(\frac{\pi}{4}\right)\right)=\sqrt{2} \sin \left(x-\frac{\pi}{4}\right)$

On simplification we get

$\Rightarrow \sin x-\cos x=\sqrt{2} \sin \left(x-\frac{\pi}{4}\right)$

Now substituting these values in given equation we get

$\Rightarrow$$\lim _{x \rightarrow \frac{\pi}{4}} \frac{\sin x-\cos x}{x-\frac{\pi}{4}}=\lim _{x \rightarrow \frac{\pi}{4}} \frac{\sqrt{2} \sin \left(x-\frac{\pi}{4}\right)}{x-\frac{\pi}{4}} \Rightarrow$$\lim _{x \rightarrow \frac{\pi}{4}} \frac{\sqrt{2} \sin \left(x-\frac{\pi}{4}\right)}{x-\frac{\pi}{4}}=\sqrt{2} \lim _{x \rightarrow \frac{\pi}{4}} \frac{\sin \left(x-\frac{\pi}{4}\right)}{x-\frac{\pi}{4}}$

$\lim _{\text {Now as }} \frac{\sin x}{x}=1$

And by substituting the limits we get

$\Rightarrow$$\sqrt{2} \lim _{x \rightarrow \frac{\pi}{4}} \frac{\sin \left(x-\frac{\pi}{4}\right)}{x-\frac{\pi}{4}}=\sqrt{2} \cdot 1=\sqrt{2} \Rightarrow$$\lim _{x \rightarrow \frac{\pi}{4}} \frac{\sin x-\cos x}{x-\frac{\pi}{4}}=\sqrt{2}$