How to calculate the limit of $(\frac{x}{x+1})^x$ I am looking at the probability of losing $x$ games in a row, in a game where the probability of winning is $1/x$.  (For example, if this is a fair casino game, what is the probability of losing $x$ before winning it back)
I have calculated this to be $(\frac{x}{x+1})^x$, and I have seen in excel that as $x$ increases, this number approaches about 0.368.
I am interested in how to calculate theoretically this assymptote, particularly because I am interested in also working out the assymptote for the function $(\frac{x}{x+1})^{(x/z)}$ where $z$ is a number greater than 1. (This situation looks at unfair casino games)
Edit:
Having thought further, I have deduced my first question is $\frac{1}{e}$, however I'm not sure how to deduce my second question that involves $z$.  Thanks
 A: $$\left(\frac{x}{x+1}\right)^x=\left(1-\frac{1}{x+1}\right)^x=\left(1-\frac{1}{x+1}\right)^{x+1}\left(1-\frac{1}{x+1}\right)^{-1}$$
This tends to
$$\frac{1}{e}$$
https://en.wikipedia.org/wiki/E_%28mathematical_constant%29
For the second one, note that $\left(\frac{x}{x+1}\right)^{x/z}=\left(\left(\frac{x}{x+1}\right)^{x}\right)^{1/z}$ so this tends to $(\frac{1}{e})^{1/z}$.
A: It is well-known that
$(\frac{x+1}{x})^{x}
=(1+\frac{1}{x})^{x}
\to e
\approx 2.71828...
$
as
$x \to \infty
$.
Therefore
$(\frac{x}{x+1})^{x}
\to \frac1{e}
$
as
$x \to \infty
$.
Similarly,
$(\frac{x}{x+1})^{(x/z)}
=\left((\frac{x}{x+1})^{x}\right)^{1/z}
\to \left(\frac1{e}\right)^{1/z}
=\frac1{e^{1/z}}
$
as
$x \to \infty
$.
This assumes that $z$ is constant.
If $z \to \infty$,
all bets are off.
A: Start defining $$A=(\frac{x}{x+1})^{(x/z)}$$ Take logarithms $$\log(A)=\frac xz \log\frac{x}{x+1}=-\frac xz \log\frac{x+1}{x}=-\frac xz \log(1+\frac{1}{x})$$ Now remember that, for small values of $y$, $\log(1+y)=y-\frac{y^2}{2}+\frac{y^3}{3}+O\left(y^4\right)$; replace $y$ by $\frac{1}{x}$. So, $$\log(A)=-\frac xz \times \Big( \frac{1}{x}-\frac{1}{2 x^2}+\frac{1}{3
   x^3}+O\left(\left(\frac{1}{x}\right)^4\right) \Big)=-\frac{1}z \times(1-\frac{1}{2 x}+\frac{1}{3 x^2}+\cdots)$$ So, at the limit $$A=e^{-1/z}$$
