1. ## Limits

lim (x-->infinity) (x/ x+1)^lnx

Do I change it into an integral?

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

3. I'm a bit confused... so

I got lost after the 3rd step...and also what does e equal to?

4. Originally Posted by circuscircus
I'm a bit confused... so

I got lost after the 3rd step...and also what does e equal to?
$e = \lim_{n \to \infty} \left(1 + \frac 1n \right)^n \approx 2.71828183$

but that's not what red_dog used here. he used the fact that $\frac 1e = \lim_{n \to \infty} \left( 1 - \frac 1n \right)^n$

thus $\lim_{n \to \infty} \left( 1 - \frac 1n \right)^{-n} = \lim_{n \to \infty} \frac 1 { \left( 1 - \frac 1n \right)^{n}} = \frac 1{ \lim_{n \to \infty} \left( 1 - \frac 1n \right)^{n}} = e$

here you could replace $n$ with $x + 1$

i hope you understand the rest, if not say so. (he used L'Hopital's rule to evaluate the limit in the power)

5. Originally Posted by circuscircus
I'm a bit confused... so

I got lost after the 3rd step...and also what does e equal to?
How could you be expected to do this limit if you don't know the definition of e?? (Seeing as ln(x) is the inverse function to $e^x$.)

-Dan