Question

Evaluate:
$\underset{x\to 0}{lim}\frac{1-\left(\cos x\right)\left(\sqrt{\cos 2x}\right)}{x^4}$

Answer 1

$\underset{x\to 0}{lim}\frac{1-\cos \left(x\right)\left(\sqrt{\cos 2x}\right)}{x^4}=l$ (say)

This limit is the form $\frac{0}{0}$, so apply L' Hospital Rule,

$l=\underset{x\to 0}{lim}\frac{-\cos x[\frac{1}{2\sqrt{\cos 2x}}.(-\sin 2x).\left(2\right)]+\left(\sin x\right)\sqrt{\cos 2x}}{4x^3}$

$l=\underset{x\to 0}{lim}\frac{1}{4}\left[\frac{\cos x\sin 2x+\sin x\cos 2x}{x^3\sqrt{\cos 2x}}\right]$

$l=\underset{x\to 0}{lim}\frac{1}{4}\left(\frac{1}{\sqrt{\cos 2x}}\right)\times \underset{x\to 0}{lim}\frac{\sin 3x}{x^3}$

$l=\frac{1}{4}\left(1\right)\underset{x\to 0}{lim}\frac{\sin 3x}{x^3}$

Apllying L' Hospital again

$l=\frac{1}{4}\underset{x\to 0}{lim}\frac{\cos 3x.3}{3x^2}$

Again L' Hospital Rule,

$l=\frac{1}{4}\underset{x\to 0}{lim}\frac{-\sin 3x.9}{3\left(2\right)x}$

Once again L' Hospital Rule

$l=\frac{1}{4}\underset{x\to 0}{lim}\frac{-\cos 3x.9}{3\left(2\right)}$

$l=\frac{1}{4}.\frac{[-\cos 3\left(0\right)]\left(3\right)}{2}$

$l=-\frac{3}{8}$