Limit of an Indeterminate Form

**mturner07** · February 23rd 2010, 04:07 PM

Hey guys, I have a problem that I'm having trouble with. Any help would be appreciated.

$\lim{x}\rightarrow\infty (1 + \frac{1}{x})^x$ = Indeterminate Form $1^\infty$

Let $y = (1 + \frac{1}{x})^x$ , $lny = ln(1 + \frac{1}{x})^x = xln(1 + \frac{1}{x})$

$\lim{x}\rightarrow\infty lny = \lim{x}\rightarrow\infty xln(1 + \frac{1}{x})$ = $\lim{x}\rightarrow\infty \frac{ln(1 + \frac{1}{x})}{\frac{1}{x}}$ = Indeterminate Form $\frac{0}{0}$ = (Using L'Hopital's Rule) $\lim{x}\rightarrow\infty \frac{\frac{1}{1 + \frac{1}{x}}}{\frac{-1}{x^2}}$

I'm stuck here. The limit of the top is 1 and the limit of the bottom is 0. I don't know if I did something wrong or if I just don't know what to do next. Thanks in advance.

**Aryth** · February 23rd 2010, 04:41 PM

I'm going to start here:

$\lim_{x \to \infty} \frac{\ln{\left(1 + \frac{1}{x}\right)}}{\frac{1}{x}}$

Using L'Hopitals rule, you must use the Quotient Rule, which will yield:

$\frac{\left(\ln{\left(1 + \frac{1}{x}\right)}\right)'\frac{1}{x} - \ln{\left(1 + \frac{1}{x}\right)}\left(\frac{1}{x}\right)'}{\lef t(\frac{1}{x}\right)^2}$

The result from this will yield a different form than the one you have provided.

That should lead you to the answer.

Good Luck.

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