What is the name and proof of the limit of this function? In a proof I find the following limit being used:
$$ \lim\limits_{x \to \infty} (1- \frac{a}{x})^x = e^{(-a)}, $$
where $a$ is a constant. Does this limit have a common name and where can I find a proof (or please give a proof)?
 A: $$L=\lim_{x \to \infty} (1 - \frac{a}{x})^x$$
Now use:
$$\frac{d}{dx}e^x=e^x$$
To get if:
$$y=e^x$$
Then
$$\frac{dy}{dx}=y$$
$$\int \frac{dy}{y}=\int dx$$
Define:
$$\int \frac{dy}{y} := \ln(y)+C$$
Then we have:
$$\ln (y)+c=x$$
Define:
$$\ln(e) :=1$$
If $y=e$ then $x=1$ and hence:
$$\ln (y)=x$$
$$\ln (e^x)=x$$
And,
$$e^{\ln (y)}=e^x=y$$
$$\ln (a^x)=\ln (e^{x \ln a})=x \ln a$$
Therefore,
$$\ln L=\lim_{x \to \infty} \frac{\ln (1-\frac{a}{x})}{\frac{1}{x}}$$
Now take the derivative top and bottom using what what we have above and the chain rule.
A: $$\left(1-\frac{a}{x}\right)^x$$
is not defined for general real $x$ yet before you defining $e^x$. So you need to define the function $e^y$ first.
If you define
$$e^y=\lim_{n\to \infty}\left(1+\frac{y}{n}\right)^n$$
then
$$\lim_{n\to \infty}\left(1-\frac{y}{n}\right)^n=e^{-y}$$
holds by definition and there is nothing to prove.
If you define
$$e^y=\sum_{n=0}^\infty\frac{y^n}{n!}$$
then you need to prove it. I cannot find a proof at the moment.
Here is a prove for the case $a=1$, to show that
$$\lim_{n\to \infty}\left(1+\frac{1}{n}\right)^n=e$$
Hope this may help:
Define
$$e=\sum_{n=0}^\infty\frac{1}{n!}$$
Let
$$s_n=\sum_{k=0}^n\frac{1}{k!}$$
and
$$t_n=\left(1+\frac{1}{n}\right)^n$$
By binomial theorem
$$t_n=1+1+\frac{1}{2!}\left(1-\frac{1}{n}\right)+\frac{1}{3!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)+\cdots+\frac{1}{n!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\cdots\left(1-\frac{n-1}{n}\right)$$
Hence
$$t_n\le s_n$$
so that
$$\limsup_{n\to\infty}t_n\le e$$
Next if $n \ge m$,
$$t_n \ge 1+1+\frac{1}{2!}\left(1-\frac{1}{n}\right)+\frac{1}{3!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)+\cdots+\frac{1}{m!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\cdots\left(1-\frac{m-1}{n}\right)$$
Let $n \to \infty$, keeping $m$ fixed. We get
$$\liminf_{n\to\infty}t_n \ge 1+1+\frac{1}{2!}+\cdots+\frac{1}{m!}$$
so that
$$s_m \le \liminf_{n\to \infty}t_n$$
Letting $m\to \infty$ we finally get
$$e \le \liminf_{n\to \infty}t_n$$
Therefore,
$$\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n=e$$
