# Math Help - Need help on limits......

1. ## Need help on limits......

HELP!

$\lim_{x\to 1} \frac{\sqrt[3]{x} - 1}{x-1} = ????$

HELP!

$\lim_{x\to 1} \frac{\sqrt[3]{x} - 1}{x-1} = ????$
By direct substitution we have that the limit looks like $\frac{0}{0}$ so we may use L'Hopital's rule:

$\lim_{x \to 1} \frac{\sqrt[3]{x} - 1}{x-1} = \lim_{x \to 1} \frac{ \frac{1}{3} x^{-2/3} }{1} = \frac{1}{3} 1^{-2/3} = \frac{1}{3}$

If that doesn't work for you, (this is gonna be weird!) note that we may write x - 1 as the difference between two cubes:
$x - 1 = \left ( x^{1/3} \right )^3 - 1^3$

And
$\left ( x^{1/3} \right )^3 - 1^3 = \left ( x^{1/3} - 1 \right ) \left ( x^{2/3} + x^{1/3} + 1 \right )$

So
$\lim_{x \to 1} \frac{\sqrt[3]{x} - 1}{x-1} = \lim_{x \to 1} \frac{x^{1/3} - 1}{\left ( x^{1/3} - 1 \right ) \left ( x^{2/3} + x^{1/3} + 1 \right )}$

= $\lim_{x \to 1} \frac{1}{x^{2/3} + x^{1/3} + 1}$

= $\frac{1}{1^{2/3} + 1^{1/3} + 1} = \frac{1}{3}$

-Dan

HELP!

$\lim_{x\to 1} \frac{\sqrt[3]{x} - 1}{x-1} = ????$
---
Rather, you can factor,
$x-1=(\sqrt[3]{x})^3-1^3=(\sqrt[3]{x}-1)(\sqrt[3]{x^2}+\sqrt[3]{x}+1)$

Thus,
$\lim_{x\to 1}\frac{\sqrt[3]{x}-1}{(\sqrt[3]{x}-1)(\sqrt[3]{x^2}+\sqrt[3]{x}+1)}$
Thus,
$\lim_{x\to 1}\frac{1}{\sqrt[3]{x^2}+\sqrt[3]{x}+1}$
Which is,
$\frac{1}{3}$