1.) lim as x--> infinity (x - cube root of x)

2.) lim as x-->0+ ln(2x)

Anyone know how to solve these limit equations? Thanks.

Printable View

- Sep 11th 2008, 03:49 PMsleepiieeLimits Help
1.) lim as x--> infinity (x - cube root of x)

2.) lim as x-->0+ ln(2x)

Anyone know how to solve these limit equations? Thanks. - Sep 11th 2008, 04:10 PMmr fantastic
- Sep 11th 2008, 04:18 PMskeeter
$\displaystyle x - x^{\frac{1}{3}} = x^{\frac{1}{3}}(x^{\frac{2}{3}} - 1) $

$\displaystyle \lim_{x \to \infty} x^{\frac{1}{3}}(x^{\frac{2}{3}} - 1)$

$\displaystyle \lim_{x \to \infty} x^{\frac{1}{3}} \cdot \lim_{x \to \infty} (x^{\frac{2}{3}} - 1)$

finish up

$\displaystyle \lim_{x \to 0^+} \ln(2x) $

$\displaystyle \lim_{x \to 0^+} (\ln{x} + \ln{2})$

$\displaystyle \lim_{x \to 0^+} \ln{x} + \lim_{x \to 0^+} \ln{2}$

finish up - Sep 11th 2008, 06:37 PMsleepiiee
I got the first one thanks. Still don't understand what to do next with the 2nd problem

- Sep 11th 2008, 06:39 PM11rdc11
From the right hand side ln(2x) goes to negative infinity as it approaches 0