Find $\underset{x\rightarrow 0}\lim{\frac{1-\cos{x}\sqrt{\cos{2x}}}{x\sin{x}}}$ 
Find
  $$\underset{x\rightarrow 0}\lim{\frac{1-\cos{x}\sqrt{\cos{2x}}}{x\sin{x}}}$$

My work.$$\underset{x\rightarrow 0}\lim\frac{1}{x\sin{x}}=\frac{\underset{x\rightarrow0}\lim{\;\frac{\sin{x}}{x}}}{\underset{x\rightarrow 0}\lim{\;x\sin{x}}}=\underset{x\rightarrow 0}\lim\frac{1}{x^2}$$
$$\underset{x\rightarrow 0}\lim{\frac{\cos{x}}{x\sin{x}}}=\underset{x\rightarrow 0}\lim{\frac{\sin{2x}}{2x\sin^2{x}}}=\underset{x\rightarrow 0}\lim{\frac{\sin{2x}}{2x}}\cdot\underset{x\rightarrow 0}\lim{\frac{1}{\sin^2{x}}}=\frac{1}{x^2}$$
$$\underset{x\rightarrow 0}\lim{\sqrt{\cos{2x}}}=\underset{x\rightarrow 0}\lim{\sqrt{1-2\sin^2{x}}}=\underset{x\rightarrow 0}\lim{\sqrt{1-2x^2}}$$
L' Hopital's rule:
$\underset{x\rightarrow 0}\lim{\frac{1-\cos{x}\sqrt{\cos{2x}}}{x\sin{x}}}=\underset{x\rightarrow 0}\lim{\frac{1-\sqrt{1-2x^2}}
{x^2}}=\underset{x\rightarrow 0}\lim{\frac{-4x}{x^3\sqrt{1-2x^2}}}$
What should I do next?
 A: As $x\to 0$, $x\sin x\sim x^2$. Also
$$\cos x=1-\frac{x^2}2+O(x^4),$$
$$\cos2x=1-2x^2+O(x^4),$$
$$\sqrt{\cos2x}=1-x^2+O(x^4),$$
$$(\cos x)(\sqrt{\cos2x})=1-\frac{3x^2}2+O(x^4),$$
$$1-(\cos x)(\sqrt{\cos2x})\sim\frac{3x^2}2.$$
Therefore
$$\lim_{x\to0}\frac{1-(\cos x)(\sqrt{\cos2x})}{x\sin x}=\frac32.$$
A: Note that, as $x\to 0$,
$$\begin{align}\frac{1-\cos{x}\sqrt{\cos{2x}}\pm\sqrt{\cos{2x}}}{x\sin{x}}&=
\sqrt{\cos{2x}}\cdot \frac{1-\cos{x}}{x^2}\cdot\frac{x}{\sin(x)}\\&\quad\qquad+2\cdot\frac{\sqrt{1-2\sin^2(x)}-1}{-2\sin^2(x)}\cdot\frac{\sin(x)}{x}\\
&\to1\cdot \frac{1}{2}\cdot 1+2\cdot\frac{1}{2}\cdot 1=\frac{3}{2}\end{align}$$
where we used $\cos(2x)=1-2\sin^2(x)$ and the stardard limits:
$$\lim_{t\to 0}\frac{\sin(t)}{t}=1\quad
\lim_{t\to 0}\frac{1-\cos(t)}{t^2}=\frac{1}{2}\quad
\lim_{t\to 0}\frac{\sqrt{1+t}-1}{t}=\frac{1}{2}.$$
A: \begin{eqnarray}
\mathcal L &=& \lim_{x\to0}\frac{1-\cos x \sqrt{\cos 2x}}{x\sin x}=\\
&=& \lim_{x\to0}\frac{1-\cos^2 x \cos 2x}{x\sin x (1+\cos x\sqrt{\cos 2x})}=\\
&=& \lim_{x\to0}\frac{1-\cos^2x(2\cos^2x-1)}{2x\sin x}=\\
&=& \lim_{x\to0}\frac{1-2\cos^4x+\cos^2x}{2x\sin x}=\\
&=&\lim_{x\to0}\frac{2\left(\frac12 + \cos^2 x\right)\left(1-\cos^2 x\right)}{2x\sin x}=\\
&=&\lim_{x\to0} \frac32 \frac{1-\cos^2 x}{x\sin x}=\frac32.
 \end{eqnarray}
A: By standard limits we have
$$\frac{1-\cos{x}\sqrt{\cos{2x}}}{x\sin{x}}=\frac{1-\cos x +\cos x-\cos{x}\sqrt{\cos{2x}}}{x\sin{x}}=$$
$$=\frac{1-\cos x }{x\sin{x}}+\cos x\frac{1-\sqrt{\cos{2x}}}{x\sin{x}}=$$
$$=\frac{1-\cos x }{x^2}\frac{x }{\sin{x}}+\cos x\frac{1-\sqrt{\cos{2x}}}{x\sin{x}}\frac{1+\sqrt{\cos{2x}}}{1+\sqrt{\cos{2x}}} \to \frac 32$$
indeed
$$\frac{1-\cos x }{x^2}\frac{x }{\sin{x}} \to \frac12$$
and
$$\cos x\frac{1-\sqrt{\cos{2x}}}{x\sin{x}}\frac{1+\sqrt{\cos{2x}}}{1+\sqrt{\cos{2x}}}=\frac{\cos x}{1+\sqrt{\cos{2x}}}\frac{1-\cos{2x}}{x\sin{x}}=$$
$$=\frac{\cos x}{1+\sqrt{\cos{2x}}}\frac{1-\cos{2x}}{4x^2}\frac{4x}{\sin x}\to \frac12\cdot \frac12\cdot 4=1$$
Refer also to the related


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*How to prove that $\lim\limits_{x\to0}\frac{\sin x}x=1$?

*Limits of trig functions
