How can I find limits without L'Hopital's Rule?

My question is, how can I evaluate limits without L'Hopital's Rule ?

$$\lim\limits_{x \to 0} \frac{\sin(\sqrt[3]{x})}{1-\cos x}$$

Answer

For small $x$, $1-\cos x= 2\sin^2\frac{x}{2}\approx\frac{x^2}{2}$ while $\sin\sqrt[3]{x}\approx\sqrt[3]{x}$, so $\frac{\sin\sqrt[3]{x}}{1-\cos x}\approx 2x^{-5/3}$ diverges as $x\to 0$. Thus the limit is $\infty$.