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

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


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

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 $\frac{1}{2}$:

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


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

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.


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

| cite | improve this answer | |
  • $\begingroup$ 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

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 $$\lim_{x \to 0} \frac{\sin x}{x}=1.$$

| cite | improve this answer | |
  • $\begingroup$ 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

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 $\lim_{x \to 0} \frac {\sin x}{x}=1$ :

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

| cite | improve this answer | |

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

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

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

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

| cite | improve this answer | |
  • $\begingroup$ 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

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 $$...=\lim_{x\to 0}\frac{x^2}{4x^2}=\frac{1}{4}$$

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.