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

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

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

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

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.