lim (x-->infinity) (x/ x+1)^lnx
Do I change it into an integral?
$\displaystyle \displaystyle\lim_{x\to\infty}\left(\frac{x}{x+1}\ right)^{\ln x}=\lim_{x\to\infty}\left(1-\frac{1}{x+1}\right)^{\ln x}=\lim_{x\to\infty}\left[\left(1-\frac{1}{x+1}\right)^{-(x+1)}\right]^{-\frac{ln x}{x+1}}=$
$\displaystyle \displaystyle=e^{\displaystyle-\lim_{x\to\infty}\frac{\ln x}{x+1}}=e^{\displaystyle-\lim_{x\to\infty}\frac{1}{x}}=e^0=1$
$\displaystyle e = \lim_{n \to \infty} \left(1 + \frac 1n \right)^n \approx 2.71828183$
but that's not what red_dog used here. he used the fact that $\displaystyle \frac 1e = \lim_{n \to \infty} \left( 1 - \frac 1n \right)^n$
thus $\displaystyle \lim_{n \to \infty} \left( 1 - \frac 1n \right)^{-n} = \lim_{n \to \infty} \frac 1 { \left( 1 - \frac 1n \right)^{n}} = \frac 1{ \lim_{n \to \infty} \left( 1 - \frac 1n \right)^{n}} = e$
here you could replace $\displaystyle n$ with $\displaystyle x + 1$
i hope you understand the rest, if not say so. (he used L'Hopital's rule to evaluate the limit in the power)