<s>[INST] Can be solved without L'Hopital? Can this limit be evaluated without l'hopital's rule?
$$\lim_{h\to0}\frac{\sqrt[3]{8+h}-2}{h}$$
 [/INST] \begin{align}
\frac{\sqrt[3]{8+h}-2}{h}&=
\frac{\sqrt[3]{(8+h)^2}+2\sqrt[3]{8+h}+4}{\sqrt[3]{(8+h)^2}+2\sqrt[3]{8+h}+4}
\cdot \frac{\sqrt[3]{8+h}-2}{h} \\&=\frac{(8+h)-8}{h}
\cdot\frac{1}{\sqrt[3]{(8+h)^2}+2\sqrt[3]{8+h}+4}
\to\frac{1}{12}.
\end{align}
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