Evaluating limits without L'Hopital's I need to evaluate without using L'Hopital's Rule. They say you don't appreciate something until you lose it. Turns out to be true.


*

*$\lim_{x \to 0} \frac{\sqrt[3]{1+x}-1}{x}$

*$\lim_{x \to 0} \frac{\cos 3x - \cos x}{x^2}$

I have tried to "rationalise" Q1 by using the identity $a^3-b^3=(a-b)(a^2+ab+b)$ but I still end up with a case of $\frac{0}{0}$.
For Q2, I tried to apply the formula and express $\cos 3x-\cos x =2\cos 2x \cos x$ which didn't help.
I would really be grateful for some advice on how I can proceed with the evaluating.

As regards 2), note that
$$\frac{\cos(3x) - \cos(x)}{x^2}=\frac{1 - \cos x}{x^2}-9\cdot\frac{1 - \cos (3x)}{(3x)^2}.$$
Hence
$$\lim_{x \to 0} \frac{\cos(3x) - \cos(x)}{x^2}=
\lim_{x \to 0}\frac{1 - \cos x}{x^2}-9\cdot\lim_{x \to 0}\frac{1 - \cos (3x)}{(3x)^2}\\=(1-9)\lim_{x \to 0}\frac{1 - \cos x}{x^2}
=-8\lim_{x \to 0}\frac{1 - \cos^2 x}{x^2(1+\cos x)}=-\frac{8}{2}\lim_{x \to 0}\left(\frac{\sin x}{x}\right)^2=-4.$$
