TITLE: Can be solved without L'Hopital?
QUESTION [6 upvotes]: Can this limit be evaluated without l'hopital's rule?
$$\lim_{h\to0}\frac{\sqrt[3]{8+h}-2}{h}$$

REPLY [5 votes]: One neat way to solve some limits is to use Taylor series.  If you recall the Maclaurin expansion
$$(1 + x)^n = 1 + nx + \mathcal{O}(x^2)$$
you can see that
$$\sqrt[3]{8 + h} -2 = 2\sqrt[3]{1 + h/8} -2 \approx 2\left(1 + \frac{h}{24}\right) -2 = \frac{h}{12}$$
so the limit becomes
$$\lim_{h \to 0} \frac{h/12}{h} = \frac{1}{12}.$$