2

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?

Robert Z
  • 145,942
PinkyWay
  • 4,565
  • 3
    I think you should start again, and avoid the Hospital. – Angina Seng Nov 25 '19 at 15:53
  • 2
    using series would be mush more easier, you have $\cos x\sim1-x^2/2$, $(1+x)^{1/2}\sim1+x/2$, $\sin x\sim x$ – Nanayajitzuki Nov 25 '19 at 15:57
  • 1
    Though your identities are technically correct, they are as such irrelevant to the resolution. And your application of L'Hospital's rule is invalid. –  Nov 25 '19 at 16:00

4 Answers4

3

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 $$\lim_{x\to0}\frac{1-(\cos x)(\sqrt{\cos2x})}{x\sin x}=\frac32.$$

Angina Seng
  • 158,341
3

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: $$\lim_{t\to 0}\frac{\sin(t)}{t}=1\quad \lim_{t\to 0}\frac{1-\cos(t)}{t^2}=\frac{1}{2}\quad \lim_{t\to 0}\frac{\sqrt{1+t}-1}{t}=\frac{1}{2}.$$

Robert Z
  • 145,942
3

\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}

dfnu
  • 7,528
2

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

user
  • 154,566