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

Revised to avoid l’Hospital’s rule:

Your second one can be finished off like this:

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

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 $(a^3-b^3)=(a-b)(a^2+ab+b^2)$.

For the first one, you need to rationalize. The formula for $a^3-b^3$ yields:

$$(\sqrt[3]{1+x}-1)(\sqrt[3]{(1+x)^2}+\sqrt[3]{1+x}+1)= (1+x)-1 \,.$$

Alternately, what you have there is the definition of the derivative of $\sqrt[3]{1+x}$ at $x=0$.

For the second one, at this step you are done:

$$\lim_{x\to 0}\frac{-2\sin2x*\sin x}{x^2} \,.$$

Just use

$$\lim_{x\to 0}\frac{\sin2x}{2x}=\lim_{x\to 0}\frac{\sin x}{x}=1$$