Can be solved without L'Hopital? Can this limit be evaluated without l'hopital's rule?
$$\lim_{h\to0}\frac{\sqrt[3]{8+h}-2}{h}$$
 A: 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
$$\lim_{h \to 0} \frac{h/12}{h} = \frac{1}{12}.$$
A: \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}
A: Try to let: $u= (8 + h)^{1/3} $
Thus $u^3 = 8 + h$
$h = u^3 - 8$
Then the limit becomes:
$$\lim_{u\to2}\frac{u-2}{u^3-8} = \lim_{u\to2}\frac{u-2}{(u-2)(u^2+2u+4)}=\lim_{u\to2}\frac{1}{u^2+2u+4}$$
A: $$x^3-y^3=(x-y)(x^2+xy+y^2)\implies x-y=\frac{x^3-y^3}{x^2+xy+y2}$$
Now we put
$$x=\sqrt[3]{8+h}\;,\;\;y=2\implies \sqrt[3]{8+h}-2=\frac{8+h-8}{(8+y)^{2/3}+2\sqrt[3]{8+h}+4}\implies$$
$$\frac{\sqrt[3]{8+h}-2}h=\frac1{(8+h)^{2/3}+2\sqrt[3]{8+h}+4}\xrightarrow[h\to 0]{}\frac1{8^{2/3}+2\sqrt[3]8+4}=\ldots$$ 
A: HINT:
Think about the definition of a derivative, recognize this expression as the derivative of something. (And yes, this can be found without l'Hopital's rule).
