Question

Solve $\underset{x\to \frac{\pi }{4}}{lim}\frac{\sqrt{2}\cos x-1}{\cot x-1}$

Answer 1

We have,

$\underset{x\to \frac{\pi }{4}}{lim}\left(\frac{\sqrt{2}\mathrm{cosx}-1}{\cot x-1}\right)$

This is $\frac{0}{0}$form.

So, apply L-Hospital rule

$\underset{x\to \frac{\pi }{4}}{lim}\left(\frac{-\sqrt{2}\mathrm{sinx}-0}{-\cos e{c}^{2}x-0}\right)$

$\underset{x\to \frac{\pi }{4}}{lim}\left(\frac{\sqrt{2}\mathrm{sinx}}{\cos e{c}^{2}x}\right)$

$\underset{x\to \frac{\pi }{4}}{lim}\left(\sqrt{2}{\sin }^{3}x\right)$

$=\sqrt{2}\times {\left(\sin \frac{\pi }{4}\right)}^3$

$=\sqrt{2}\times {\left(\frac{1}{\sqrt{2}}\right)}^3$

$=\frac{1}{2}$

Hence, the value is $\frac{1}{2}$.