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I've been trying to solve this over and over without L'Hopital but keep on failing:

$$\lim_{x\to0}\frac{1-\sqrt{\cos x}}{x^2}$$


My first attempt involved rationalizing:

$$\frac{1-\sqrt{\cos x}}{x^2} \cdot \frac{1+\sqrt{\cos x}}{1+\sqrt{\cos x}} = \frac{1-\cos x}{x\cdot x \cdot (1+\sqrt{\cos x})}$$

Using the rule $\frac{1-\cos x}{x} = 0$ for $x\to0$ is useless because we would end up with

$$\frac{0}{0\cdot x \cdot \sqrt{\cos x}} = \frac{0}{0}$$

But hey, perhaps we can rationalize again?

$$\frac{1-\cos x}{(x^2+x^2\sqrt{\cos x})}\cdot \frac{(x^2-x^2\sqrt{\cos x})}{(x^2-x^2\sqrt{\cos x})}$$

Resulting in

$$\frac{(1-\cos x)(x^2-x^2\sqrt{\cos x})}{x^4 - x^4\cdot\cos x} = \frac{(1-\cos x)(x^2-x^2\sqrt{\cos x})}{x^4 \cdot (1-\cos x)}$$

Cancelling

$$\frac{(x^2-x^2\sqrt{\cos x})}{x^4} = \frac{1-\sqrt{\cos x}}{x^2}$$

Well that was hilarious. I ended up at the beginning! Dammit.


My second attempt was to use the definition $\cos x = 1 - 2\sin \frac{x}{2}$:

$$\lim_{x\to0}\frac{1-\sqrt{\cos x}}{x^2} = \frac{1-\sqrt{1 - 2\sin \frac{x}{2}}}{x^2}$$

And then rationalize

$$\frac{1-\sqrt{1 - 2\sin \frac{x}{2}}}{x^2} \cdot \frac{1+\sqrt{1 - 2\sin \frac{x}{2}}}{1+\sqrt{1 - 2\sin \frac{x}{2}}}$$

$$\frac{1-(1 - 2\sin \frac{x}{2})}{x^2+x^2\sqrt{1 - \sin \frac{x}{2}}} = \frac{- 2\sin \frac{x}{2}}{x^2\left(1+\sqrt{1 - \sin \frac{x}{2}}\right)}$$

I want to make use of the fact that $\frac{\sin x}{x} = 1$ for $x\to0$, so I will multiply both the numerator and denominator with $\frac{1}{2}$:

$$\frac{-\sin \frac{x}{2}}{\frac{x}{2}\cdot x\left(1+\sqrt{1 - \sin \frac{x}{2}}\right)}$$

Then

$$\frac{-1}{x\left(1+\sqrt{1 - \sin \frac{x}{2}}\right)}$$

Well clearly that's not gonna work either. I will still get $0$ in the denominator.


The correct answer is $\frac{1}{4}$. I can kind of see why is the numerator $1$, but no idea where is that $4$ going to come out of.

I don't know how am I supposed to solve this without L'Hopital.

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HINT: $$\frac{1-\sqrt{\cos(x)}}{x^2}=\frac{1-\cos(x)}{x^2(1+\sqrt{\cos(x)})}=\frac{1-\cos(x)^2}{x^2(1+\sqrt{\cos(x)})(1+\cos(x))}$$

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  • $\begingroup$ I don't really see how does multiplying by $\frac{(1+\cos(x))}{(1+\cos(x))}$ help me. $\endgroup$ – Zol Tun Kul Sep 27 '15 at 9:37
  • $\begingroup$ but i see it HINT $$\frac{1-\cos(x)^2}{x^2}=\frac{\sin(x)^2}{x^2}$$ $\endgroup$ – Dr. Sonnhard Graubner Sep 27 '15 at 9:39
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You are supposed to use the limit $$ \lim_{x \to 0} \frac{1-\cos x}{x^2}=\frac12. $$ This should be known to you as soon as you know that $$\lim_{x \to 0} \frac{\sin x}{x}=1.$$

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  • $\begingroup$ Hm. I do know that $\frac{\sin x}{x} = 1$, but I am not quite sure why does that imply that $\frac{1-\cos x}{x^2} = \frac{1}{2}$. $\endgroup$ – Zol Tun Kul Sep 27 '15 at 9:26
  • $\begingroup$ Multiply and divide by $1+\cos x$ and recall that $1-\cos^2 x = \sin^2 x$. $\endgroup$ – Siminore Sep 27 '15 at 9:38
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After your first rationalization, you showed that the original expression is equivalent to

$$\lim_{x\to 0} \frac{1-\cos x}{x^2(1+\sqrt{\cos x})}=\lim_{x \to 0}\color{green}{\frac{1-\cos x}{x^2}} \cdot \color{blue}{\frac {1}{1+\sqrt{\cos x}}}$$

Since the $\color{green}{\text{green}}$ part of the limit converges to $\frac 12$ (it's a notable limit) and the $\color{blue}{\text{blue}}$ coloured function of $x$ is continuous in a neighborhood of $0$, we can conclude that the limit exists and its value is

$$ \lim_{x \to 0} \frac12 \cdot \frac{1}{1+\sqrt{\cos x}}= \frac 12 \cdot \frac{1}{1+\sqrt{\cos(0)}}=\frac 12 \cdot \frac12= \color{red}{\frac14}$$

Note: in case you did not know the aforementioned notable limit, here is a quick proof of it using the fact that $\lim_{x \to 0} \frac {\sin x}{x}=1$ :

$$ \lim_{x\to 0} \frac{1- \cos x}{x^2}$$ $$ = \lim_{x\to 0} \frac{1- \cos x}{x^2} \cdot \color{red}{\frac{1+ \cos x}{1+\cos x}}$$ $$ = \lim_{x \to 0} \frac{\sin^2 x}{x^2}\cdot \frac{1}{1+\cos x}= \frac 12$$

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If L'Hopitals' Rule gives you $\frac{f'}{g'}\rightarrow\frac{0}{0}$, apply L'Hopital's Rule to it. :-)

$$\frac{f(x)}{g(x)}=\frac{1-\sqrt{cos(x)}}{x^{2}} $$

$$f''(x)=-\frac{1}{2}\sqrt{cos(x)}-\frac{sin^{2}(x)}{4cos^{\frac{3}{2}}(x)} $$

$$\frac{f''(x)}{g''(x)}\rightarrow-\frac{1}{4} $$

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  • $\begingroup$ Someone posted a problem here once where you had to calculate the SIXTH derivatives before they didn't both go to zero. $\endgroup$ – Jerry Guern Sep 27 '15 at 9:39
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May be you can use the fact that $$\sqrt{1+x}=1+\frac{x}{2}+o(x)$$ and $$\cos(x)=1-\frac{x^2}{2}+o(x^2).$$ Therefore $$...=\lim_{x\to 0}\frac{x^2}{4x^2}=\frac{1}{4}$$

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  • $\begingroup$ What is $o(x)$? $\endgroup$ – Zol Tun Kul Sep 27 '15 at 9:38

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