Can be solved without L'Hopital?

Question

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

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

Answer 1

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

Now we put

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

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

Answer 2

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).

Answer 3

\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}

Answer 4

Try to let: $u= (8 + h)^{1/3} $

Thus $u^3 = 8 + h$

$h = u^3 - 8$

Then the limit becomes:

$$\lim_{u\to2}\frac{u-2}{u^3-8} = \lim_{u\to2}\frac{u-2}{(u-2)(u^2+2u+4)}=\lim_{u\to2}\frac{1}{u^2+2u+4}$$

Answer 5

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 $$\lim_{h \to 0} \frac{h/12}{h} = \frac{1}{12}.$$