I'm having trouble evaluating this limit as $x$ approaches $0$ without using Hôpital's rule.
$$\lim_{x \to 0} \frac{1-\cos^4(x)}{x^2}$$
I'd appreciate any hints on how to proceed.
EDIT: I totally forgot that
$\lim_{x \to 0} \frac{\sin(x)}{x} = 1$
I shall write my solution as punishment:
$$\lim_{x \to 0} \frac{1-\cos^4(x)}{x^2}=\lim_{x \to 0} \frac{(1-\cos^2(x))\cdot(1+\cos^2(x))}{x^2}=$$ $$=\lim_{x \to 0} \frac{1-\cos^2(x)}{x^2} \cdot \lim_{x \to 0} 1+\cos^2(x)=$$ $$=\lim_{x \to 0}\Bigl(\frac{\sin(x)}{x}\Bigr)^2\cdot2=1\cdot2=$$ $$=2$$
Hint: We have
$$1-\cos^{4}x = \left(1+\cos^2 x\right)\left(1-\cos^{2}x\right) .$$