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.

Since $\cos(x) \sim -x^2/2$ if $x \to 0$, the second argument of your second limit is $\frac{-9x^2/2 + x^2/2}{x^2}$ which evaluates to $-4$; as for your first limit, since $(x+1)^{a} -1 \sim ax$ if $a > 0$ and $x\to 0$ you get that the argument is $(1/3)x /x$ which evaluates to $1/3$.
Note that $\sim$ means asymptotic to. Also, you might say that these are the first terms of the Taylor series of the functions in your limit, but those asymptotics can be proven without Taylor series.
