Solving $\lim_{x\to0}\frac{1-\sqrt{\cos x}}{x^2}$ 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.
 A: 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 $$\lim_{x \to 0} \frac{\sin x}{x}=1.$$
A: HINT: $$\frac{1-\sqrt{\cos(x)}}{x^2}=\frac{1-\cos(x)}{x^2(1+\sqrt{\cos(x)})}=\frac{1-\cos(x)^2}{x^2(1+\sqrt{\cos(x)})(1+\cos(x))}$$
A: 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 $\lim_{x \to 0} \frac {\sin x}{x}=1$ :
$$ \lim_{x\to 0} \frac{1- \cos x}{x^2}$$
$$ = \lim_{x\to 0} \frac{1- \cos x}{x^2} \cdot \color{red}{\frac{1+ \cos x}{1+\cos x}}$$
$$ = \lim_{x \to 0} \frac{\sin^2 x}{x^2}\cdot \frac{1}{1+\cos x}= \frac 12$$
A: 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
$$...=\lim_{x\to 0}\frac{x^2}{4x^2}=\frac{1}{4}$$
A: If L'Hopitals' Rule gives you $\frac{f'}{g'}\rightarrow\frac{0}{0}$, apply L'Hopital's Rule to it.    :-)
$$\frac{f(x)}{g(x)}=\frac{1-\sqrt{cos(x)}}{x^{2}}
 $$
$$f''(x)=-\frac{1}{2}\sqrt{cos(x)}-\frac{sin^{2}(x)}{4cos^{\frac{3}{2}}(x)}
 $$
$$\frac{f''(x)}{g''(x)}\rightarrow-\frac{1}{4}
 $$
