calculus - Can be solved without L'Hopital?

Can this limit be evaluated without l'hopital's rule?

$$\lim_{h\to0}\frac{\sqrt[3]{8+h}-2}{h}$$

Answer

$$x^3-y^3=(x-y)(x^2+xy+y^2)\implies x-y=\frac{x^3-y^3}{x^2+xy+y2}$$

Now we put

$$x=\sqrt[3]{8+h}\;,\;\;y=2\implies \sqrt[3]{8+h}-2=\frac{8+h-8}{(8+y)^{2/3}+2\sqrt[3]{8+h}+4}\implies$$

$$\frac{\sqrt[3]{8+h}-2}h=\frac1{(8+h)^{2/3}+2\sqrt[3]{8+h}+4}\xrightarrow[h\to 0]{}\frac1{8^{2/3}+2\sqrt[3]8+4}=\ldots$$