limit of $ f(n) = 100 \left (1 - \frac{1}{n}\right) ^{ n}$ So I was daydreaming about math (like I do frequently) and I came up with this question/riddle.
Say you have a die. If you roll a 1 you lose, otherwise, you win. This die has n sides on it, and you roll the die n times. Are your odds better with a large n, or a small n?
If figured out that the function to calculate your odds of winning (with n sides and n rolls) is 
$$ f(n) = 100 \left (1 - \frac{1}{n}\right) ^{ n}$$
Using this function, I found out the larger the n, the better your odds of not rolling a 1. I made a short python program here to try and see what it converges upon. It seems to converge upon 36.7877 ish, but how do I find out what the exact limit is?
I think that this might be an important number, and it reminds me of this question. I'm not claiming that this would be as important as e, but it seems like it should be important.
TL;DR What is the limit of 
$\displaystyle f(n) = 100 \left (1 - \frac{1}{n}\right) ^{ n}$, as n approaches infinity?
 A: It is a well-known result that:
$$
\lim_{n \to \infty} \left(1 + \frac{a}{n}\right)^n = e^{a}
$$
In this case, $a = -1$ and $e^{-1} \approx 0.367879441\ldots$
Derivation.
Take the log of the equation, we get:
$$
\lim_{n \to \infty} \left(1 + \frac{a}{n}\right)^n = e^{\lim_{n \to \infty}n\log(1+\frac{a}{n})}
$$
Restate the exponent as:
$$
\lim_{n \to \infty}\frac{\log(1+\frac{a}{n})}{\frac{1}{n}}
$$
Now we have an indeterminate form of $\frac{0}{0}$ and can apply L'hopital:
$$
\lim_{n \to \infty}\frac{\frac{1}{1 + \frac{a}{n}}\cdot a \frac{-1}{n^2}}{\frac{-1}{n^2}}
$$
The powers of $n$ go to infinity at exactly the same rate, so they cancel, leaving:
$$
\lim_{n \to \infty}\frac{1}{1 + \frac{a}{n}}\cdot a
$$
Which is just $a$, so the original limit is $e^a$.
A: The $\lim_{n\to\infty}(1-\frac{1}{n})^{n}=e^{-1}$ so the overall $\lim_{n\to\infty}100(1-\frac{1}{n})^{n}=100e^{-1}\approx 36.7879$.
A: $$\begin{align*}
\lim_{n\to+\infty}100\left(1-\frac{1}{n}\right)^n&=
\lim_{n\to+\infty}100\left(1+\frac{1}{-n}\right)^n\\
&=\lim_{n\to+\infty}100\left[\left(1+\frac{1}{-n}\right)^{-n}\right]^{-1}\\
&=100e^{-1}\\
&=\frac{100}{e}.
\end{align*}
$$
