# Limits

• Feb 27th 2011, 09:42 PM
ewqewq
Limits
I am having trouble with this limit, and I would appreciate anyone's imput.

limit (1+(1/3X))^(7+2X)
as x->0

I have tried to manipulate this into the 2 indeterminate forms that work with L'hopitals rule to no avail.
I am leaning toward the limit approaching 0 , since the limit is undefined at x=0, if we approach x from 0+, the limit goes to infinity^7 , which means 1/[(inifinity)^-7)] = 0

• Feb 27th 2011, 09:56 PM
Prove It
Is this $\displaystyle \lim_{x \to 0}\left[1 + \left(\frac{1}{3x}\right)^{7 + 2x}\right]$ or $\displaystyle \lim_{x \to 0}\left[\left(1 + \frac{1}{3x}\right)^{7+2x}\right]$, or some other form I haven't written?
• Feb 27th 2011, 09:59 PM
ewqewq
It is the second one . I am sorry mine was not clear.
• Feb 27th 2011, 10:17 PM
Prove It
$\displaystyle \lim_{x \to 0}\left[\left(1 + \frac{1}{3x}\right)^{7+2x}\right] = \lim_{x \to 0}\left\{e^{\ln{\left[\left(1+\frac{1}{3x}\right)^{7+2x}\right]}}\right\}$

$\displaystyle = \lim_{x \to 0}\left[e^{(7+2x)\ln{\left(1 + \frac{1}{3x}\right)}}\right]$

$\displaystyle = e^{\lim_{x \to 0}\left[(7+2x)\ln{\left(1 + \frac{1}{3x}\right)}\right]}$.

It doesn't look like you need to use L'Hospital's Rule at all...
• Feb 27th 2011, 10:28 PM
ewqewq
Thanks for your help, however I still do not understand what would happen to 1/(3x), since it would be undefined