# Thread: 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

$\displaystyle \lim_{h\to 0}$ $\displaystyle \frac{{\sqrt[3]{{8 + h}} - 2}}{h}$$\displaystyle \Rightarrow \lim_{h \to 0} \frac{{\frac{1}{{3\sqrt[3]{{\left( {8 + h} \right)^2 }}}}}}{1}$$\displaystyle \Rightarrow \frac{1}{{3\sqrt[3]{{\left( {8 + 0} \right)^2 }}}} \Rightarrow \frac{1}{{3\sqrt[3]{{64}}}}$ $\displaystyle \Rightarrow \frac{{\sqrt[3]{{64^2 }}}}{{192}}$ $\displaystyle \Rightarrow \frac{{\sqrt[3]{{4096}}}}{{192}} \Rightarrow \frac{{16}}{{192}}$$\displaystyle \Rightarrow \underline{\underline {{\rm{ }}\frac{1}{{12}}{\rm{ }}}} Mmm Latex is strange ^^ 6. Originally Posted by sodk lim (cube rt (8+h) - 2) / h h->0 expanding on Ted's excellent hint ... \displaystyle f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} 7. Hello, sodk! Here is Ted's suggestion . . . in baby-steps. Recall that: .\displaystyle (a-b)(a^2 + ab + b^2)\;=\;a^3-b^3 Find: .\displaystyle \lim_{h\to0} \frac{\sqrt[3]{8+h} - 2}{h} Multiply by: .\displaystyle \frac{\sqrt[3]{(8+h)^2} + 2\sqrt[3]{8+h} + 4} {\sqrt[3]{(8+h)^2} + 2\sqrt[3]{8+h} + 4} \displaystyle \frac{\sqrt[3]{8+h} - 2}{h} \cdot\frac{\sqrt[3]{(8+h)^2} + 2\sqrt[3]{8+h} + 4} {\sqrt[3]{(8+h)^2} + 2\sqrt[3]{8+h} + 4} \;\;=\;\;\frac{(8+h) - 8} {h\left[\sqrt[3]{(8+h)^2} + 2\sqrt[3]{8+h} + 4\right]} . . . \displaystyle =\;\;\frac{h} {h\left[\sqrt[3]{(8+h)^2} + 2\sqrt[3]{8+h} + 4\right]} \;\;=\;\;\frac{1}{\sqrt[3]{(8+h)^2} + 2\sqrt[3]{8+h} + 4} Therefore: .\displaystyle \lim_{h\to0}\left[ \frac{1}{\sqrt[3]{(8+h)^2} + 2\sqrt[3]{8+h} + 4}\right]\;\;=\;\;\frac{1}{(\sqrt[3]{8})^2 + 2\sqrt[3]{8} + 4}\;\;=\;\;\frac{1}{4+4+4}\;\;=\;\;\frac{1}{12} 8. match up ... \displaystyle f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} to \displaystyle f'(8) = \lim_{h \to 0} \frac{\sqrt[3]{8+h} - \sqrt[3]{8}}{h} and realize that \displaystyle f(x) = \sqrt[3]{x} \displaystyle f'(x) = \frac{1}{3\sqrt[3]{x^2}} and the desired limit is the value of \displaystyle f'(8) ... \displaystyle f'(8) = \frac{1}{12} 9. Originally Posted by Soroban Hello, sodk! Here is Ted's suggestion . . . in baby-steps. Recall that: .\displaystyle (a-b)(a^2 + ab + b^2)\;=\;a^3-b^3 Multiply by: .\displaystyle \frac{\sqrt[3]{(8+h)^2} + 2\sqrt[3]{8+h} + 4} {\sqrt[3]{(8+h)^2} + 2\sqrt[3]{8+h} + 4} \displaystyle \frac{\sqrt[3]{8+h} - 2}{h} \cdot\frac{\sqrt[3]{(8+h)^2} + 2\sqrt[3]{8+h} + 4} {\sqrt[3]{(8+h)^2} + 2\sqrt[3]{8+h} + 4} \;\;=\;\;\frac{(8+h) - 8}$$\displaystyle {h\left[\sqrt[3]{(8+h)^2} + 2\sqrt[3]{8+h} + 4\right]}$

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

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