Find the following limit $\lim_{x\to 0}\frac{\sqrt[3]{1+x}-1}{x}$ and $\lim_{x\to 0}\frac{\cos 3x-\cos x}{x^2}$ 
Find the following limits
$$\lim_{x\to 0}\frac{\sqrt[3]{1+x}-1}{x}$$

Any hints/solutions how to approach this? I tried many ways, rationalization, taking out x, etc. But I still can't rid myself of the singularity. Thanks in advance.
Also another question.

Find the limit of
$$\lim_{x\to 0}\frac{\cos 3x-\cos x}{x^2}$$

I worked up till here, after which I got stuck. I think I need to apply the squeeze theore, but I am not sure how to.
$$\lim_{x\to 0}\frac{\cos 3x-\cos x}{x^2} = \lim_{x\to 0}\frac{-2\sin\frac{1}{2}(3x+x)\sin\frac{1}{2}(3x-x)}{x^2}=\lim_{x\to 0}\frac{-2\sin2x\sin x}{x^2}=\lim_{x\to 0}\frac{-2(2\sin x\cos x)\sin x}{x^2}=\lim_{x\to 0}\frac{-4\sin^2 x\cos x}{x^2}$$
Solutions or hints will be appreciated. Thanks in advance! L'hospital's rule not allowed.

Like N.S. said, looking this limit as derivative is a way to solve.
You could also do $u=x+1$ to simplify your expression and consider $f(u)=u^{1/3}$.
$$u=x+1\rightarrow \lim_{u\rightarrow 1} \frac{u^{1/3}-1}{u-1}=\lim_{u\rightarrow 1} \frac{f(u)-f(1)}{u-1}=f'(1)$$
But $f'(u) = \frac{1}{3}u^{-2/3}$, then $f'(1) = \frac{1}{3}\cdot 1^{-2/3}=\frac{1}{3}$.
