Question

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

• A. $\frac{1}{16\sqrt{2}}$
• B. $\frac{1}{32\sqrt{2}}$
• C. $\frac{1}{16}$
• D. $-\frac{1}{2\sqrt{2}}$

Answer 1

The correct option is D $-\frac{1}{2\sqrt{2}}$

$\underset{x\to \frac{\pi }{4}}{lim}\frac{\cos x-\sin x}{(4x-\pi )}$ $\left(\frac{0}{0}\text{Form}\right)$
Use L-Hospital's rule

$\underset{x\to \frac{\pi }{4}}{lim}\frac{-\sin x-\cos x}{4}$
Now, the $\frac{0}{0}$ form is removed, so we can directly substitute the value of $x=\frac{\pi }{4}$
So, the limit $=\frac{-\sqrt{2}}{4}=-\frac{1}{2\sqrt{2}}$