How to calculate $ \lim_{x\to\infty} (\frac{x}{x+1})^x$ using L'Hopitals rule? I am trying to calculate $ \lim_{x\to\infty} (\frac{x}{x+1})^x$ using L'Hopital.

Apparently without L'Hopital the limit is
$$ \lim_{x\to\infty} (\frac{x}{x+1})^x = \lim_{x\to\infty} (1 + \frac{-1}{x+1})^x = \lim_{x\to\infty} (1 - \frac{1}{x+1})^{x+1} \frac{1}{1-\frac{1}{x+1}} = e^{-1} * \frac{1}{1} = \frac{1}{e}$$
I am wondering how one could calculate this limit using L'Hopital's rule.

My failed approach
My initial syllogism was to use the explonential-log trick in combination with the chain rule as following:
$$ \lim_{x\to\infty} (\frac{x}{x+1})^x = e^{\lim_{x\to\infty} x \ln(\frac{x}{x+1})} \quad (1) $$
So, basically the problem that way is reduced to:
$$ \lim_{x\to\infty} x \ln(\frac{x}{x+1}) = \lim_{x\to\infty} x * \lim_{x\to\infty}\ln(\frac{x}{x+1}) \quad (2)$$
As far as $\ln(\frac{x}{x+1})$ is concerned, it has the form $f(g(x))$, so using the chain rule for limits and chain rule for derivatives in order to apply L'Hopital we can rewrite it as:
$$ \lim_{x\to\infty} \ln( \lim_{x\to\infty} \frac{(x)'}{(x+1)'}) =   \lim_{x\to\infty}  ln(1)  \quad (3)$$
But $(2),(3) \to 0 * \infty$, so that failed.
 
Any ideas on how we could approach this in other ways?
 A: You wrote:
"$\lim_{x\to\infty} x \ln(\frac{x}{x+1}) = \lim_{x\to\infty} x * \lim_{x\to\infty}\ln(\frac{x}{x+1}).$"
Observe that $\lim_{x\to\infty} x = \infty $ !
For $\lim_{x\to\infty} x \ln(\frac{x}{x+1})$ let $f(x):= \ln(\frac{x}{x+1})$ and $g(x):=1/x.$ Then we have
$$\lim_{x\to\infty} x \ln(\frac{x}{x+1})= \lim_{x\to\infty} \frac{f(x)}{g(x)}.$$
This is a limit of the form "$ \frac{0}{0}$".
Now compute $\lim_{x\to\infty} \frac{f'(x)}{g'(x)}.$
A: $x \ln (\frac x {x+1})=\frac {\ln x -\ln (x+1)} {1/x}$. So we get $\lim \frac {1/x-1/(1+x)} {-1/x^{2}}=-\lim \frac x {1+x}=-1$. Hence the answer is $e^{-1}$.
A: Caution,
$$\lim fg=\lim f\lim g$$
can only be used when the limits on the right both exist, which is not the case here.
By L'Hospital
$$\lim_{x\to\infty}\log\left(\frac x{x+1}\right)^x=\lim_{x\to\infty}\frac{\log\left(\dfrac x{x+1}\right)}{\dfrac1x}=\lim_{x\to\infty}\frac{\dfrac1x-\dfrac1{x+1}}{-\dfrac1{x^2}}=-\lim_{x\to\infty}\frac x{x+1}=-1.$$

The simplest is, by continuity of the inverse function,
$$\lim_{x\to\infty}\left(\frac x{x+1}\right)^x=\frac1{\lim_{x\to\infty}\left(1+\dfrac1x\right)^x}=\frac1e.$$
