# Math Help - How to find this limit?

1. ## How to find this limit?

lim (cube rt (8+h) - 2) / h
h->0

2. Think about the definition of the derivative ..

3. It's supposed to be solved without using L'Hopital.
Maybe using rationalization, I don't know...

4. Originally Posted by sodk
It's supposed to be solved without using L'Hopital.
Maybe using rationalization, I don't know...
Did you see post #2 ?

5. Or just use good old L`lôpital rule

$\lim_{h\to 0}$ $\frac{{\sqrt[3]{{8 + h}} - 2}}{h}$ $\Rightarrow \lim_{h \to 0} \frac{{\frac{1}{{3\sqrt[3]{{\left( {8 + h} \right)^2 }}}}}}{1}$ $\Rightarrow \frac{1}{{3\sqrt[3]{{\left( {8 + 0} \right)^2 }}}} \Rightarrow \frac{1}{{3\sqrt[3]{{64}}}}$ $\Rightarrow \frac{{\sqrt[3]{{64^2 }}}}{{192}}$ $\Rightarrow \frac{{\sqrt[3]{{4096}}}}{{192}} \Rightarrow \frac{{16}}{{192}}$ $\Rightarrow \underline{\underline {{\rm{ }}\frac{1}{{12}}{\rm{ }}}}$

Mmm Latex is strange ^^

6. Originally Posted by sodk
lim (cube rt (8+h) - 2) / h
h->0

expanding on Ted's excellent hint ...

$
f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}
$

7. Hello, sodk!

Here is Ted's suggestion . . . in baby-steps.

Recall that: . $(a-b)(a^2 + ab + b^2)\;=\;a^3-b^3$

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

Multiply by: . $\frac{\sqrt[3]{(8+h)^2} + 2\sqrt[3]{8+h} + 4} {\sqrt[3]{(8+h)^2} + 2\sqrt[3]{8+h} + 4}$

$\frac{\sqrt[3]{8+h} - 2}{h} \cdot\frac{\sqrt[3]{(8+h)^2} + 2\sqrt[3]{8+h} + 4} {\sqrt[3]{(8+h)^2} + 2\sqrt[3]{8+h} + 4} \;\;=\;\;\frac{(8+h) - 8}
{h\left[\sqrt[3]{(8+h)^2} + 2\sqrt[3]{8+h} + 4\right]}$

. . . $=\;\;\frac{h} {h\left[\sqrt[3]{(8+h)^2} + 2\sqrt[3]{8+h} + 4\right]} \;\;=\;\;\frac{1}{\sqrt[3]{(8+h)^2} + 2\sqrt[3]{8+h} + 4}$

Therefore: . $\lim_{h\to0}\left[ \frac{1}{\sqrt[3]{(8+h)^2} + 2\sqrt[3]{8+h} + 4}\right]\;\;=\;\;\frac{1}{(\sqrt[3]{8})^2 + 2\sqrt[3]{8} + 4}\;\;=\;\;\frac{1}{4+4+4}\;\;=\;\;\frac{1}{12}$

8. match up ...

$
f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}
$

to

$f'(8) = \lim_{h \to 0} \frac{\sqrt[3]{8+h} - \sqrt[3]{8}}{h}
$

and realize that $f(x) = \sqrt[3]{x}$

$f'(x) = \frac{1}{3\sqrt[3]{x^2}}$

and the desired limit is the value of $f'(8)$ ...

$f'(8) = \frac{1}{12}$

9. Originally Posted by Soroban
Hello, sodk!

Here is Ted's suggestion . . . in baby-steps.

Recall that: . $(a-b)(a^2 + ab + b^2)\;=\;a^3-b^3$

Multiply by: . $\frac{\sqrt[3]{(8+h)^2} + 2\sqrt[3]{8+h} + 4} {\sqrt[3]{(8+h)^2} + 2\sqrt[3]{8+h} + 4}$

$\frac{\sqrt[3]{8+h} - 2}{h} \cdot\frac{\sqrt[3]{(8+h)^2} + 2\sqrt[3]{8+h} + 4} {\sqrt[3]{(8+h)^2} + 2\sqrt[3]{8+h} + 4} \;\;=\;\;\frac{(8+h) - 8}$ $
{h\left[\sqrt[3]{(8+h)^2} + 2\sqrt[3]{8+h} + 4\right]}" alt="
{h\left[\sqrt[3]{(8+h)^2} + 2\sqrt[3]{8+h} + 4\right]}" />

. . . $=\;\;\frac{h} {h\left[\sqrt[3]{(8+h)^2} + 2\sqrt[3]{8+h} + 4\right]} \;\;=\;\;\frac{1}{\sqrt[3]{(8+h)^2} + 2\sqrt[3]{8+h} + 4}$

Therefore: . $\lim_{h\to0}\left[ \frac{1}{\sqrt[3]{(8+h)^2} + 2\sqrt[3]{8+h} + 4}\right]\;\;=\;\;\frac{1}{(\sqrt[3]{8})^2 + 2\sqrt[3]{8} + 4}\;\;=\;\;\frac{1}{4+4+4}\;\;=\;\;\frac{1}{12}$
That was exactly what I wanted! Unfortunately your post came late, I've solved it this exact way. My mistake was just trying to rationalize the numerator incorrectly. The trick is the factors $(a-b)(a^2 + ab + b^2)\;=\;a^3-b^3$
Anyway thanks.