Question

The value of the limit $\underset{x\to \frac{\pi }{4}}{\lim }\frac{\sin x-\cos x}{\frac{\pi }{4}-x}$ is

• A. $-\sqrt{2}$
• B. -1
• C. $\sqrt{2}$
• D. $\sqrt{3}$

Answer 1

The correct option is A $-\sqrt{2}$

$\underset{x\to \frac{\pi }{4}}{\lim }\frac{\sin x-\cos x}{\frac{\pi }{4}-x}\underset{x\to \frac{\pi }{4}}{\lim }\frac{\frac{1}{\sqrt{2}}\sin x-\frac{1}{\sqrt{2}}\cos x}{\frac{1}{\sqrt{2}}\left(\frac{\pi }{4}-x\right)}=\underset{x\to \frac{\pi }{4}}{\lim }\frac{\sin x\cos \frac{\pi }{4}-\sin \frac{\pi }{4}\cos x}{\frac{1}{\sqrt{2}}\left(\frac{\pi }{4}-x\right)}=\underset{x\to \frac{\pi }{4}}{\lim }\frac{\sin \left(x-\frac{\pi }{4}\right)}{\frac{1}{\sqrt{2}}\left(\frac{\pi }{4}-x\right)}=-\sqrt{2}\underset{x\to \frac{\pi }{4}}{\lim }\frac{\sin \left(x-\frac{\pi }{4}\right)}{\left(x-\frac{\pi }{4}\right)}Lety=x-\frac{\pi }{4}.Now,ifx\to \frac{\pi }{4}theny\to 0=-\sqrt{2}\underset{y\to 0}{\lim }\frac{\sin y}{y}=-\sqrt{2}$