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.

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