Question

$\underset{x\to \pi /4}{lim}\frac{\sin x-\cos x}{\pi -4x}$ is equal to:

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

Answer 1

Answer: (C)

$-\frac{1}{2\sqrt{2}}$

We have,

${lim}_{x\to \frac{\pi }{4}}\frac{\sin x-\cos x}{\pi -4x}\left(\frac{0}{0}form\right)$

Applying L’ Hospital rule and we get,

${lim}_{x\to \frac{\pi }{4}}\frac{\cos x+\sin x}{0-4}$

Taking limit and we get,

$\frac{\cos \frac{\pi }{4}+\sin \frac{\pi }{4}}{-4}$

$\Rightarrow \frac{\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}}{-4}$

$\Rightarrow \frac{\frac{2}{\sqrt{2}}}{-4}$

$\Rightarrow -\frac{2}{4\sqrt{2}}$

$\Rightarrow -\frac{1}{2\sqrt{2}}$

Hence, this is the answer.