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.
Revised to avoid l’Hospital’s rule:

Your second one can be finished off like this:

$$\begin{align*}
\lim_{x\to 0}\frac{-2\sin 2x\sin x}{x^2}&amp;=-2\left(\lim_{x\to 0}\frac{\sin 2x}x\right)\left(\lim_{x\to 0}\frac{\sin x}x\right)\\
&amp;=-4\left(\lim_{x\to 0}\frac{\sin 2x}{2x}\right)\cdot1\\
&amp;=-4\;.
\end{align*}$$

Try multiplying the fraction in your first limit by 

$$\frac{(1+x)^{2/3}+(1+x)^{1/3}+1}{(1+x)^{2/3}+(1+x)^{1/3}+1}$$

and making use of the identity $(a^3-b^3)=(a-b)(a^2+ab+b^2)$.
When $x$ tends to $0$, $(1+x)^n = nx$
Therefore, $(1+x)^{1/3} = x/3$.
The limit becomes $\lim_{x\to 0} \frac {1+x/3-1}{x} = 1/3$