How to prove that $\lim_{n\rightarrow\infty} (1-\frac{k}{n})^n = e^{-k}$ and $\lim_{n\rightarrow\infty} (1-\frac{k}{n})^k = 1$? How to prove that
$\lim_{n\rightarrow\infty} (1-\frac{k}{n})^n = e^{-k}$
and
$\lim_{n\rightarrow\infty} (1-\frac{k}{n})^k = 1$?
Any answer will be appreciated. Thanks.
 A: $$
\begin{array}{rclcl}
\left(1 - {k \over n}\right)^{n}
& = &
{\rm e}^{n\ln\left(1\ -\ k/n\right)}
=
{\rm e}^{n\left(-\ k/n\ -\ k^{2}/2n^{2} - \cdots\right)}
& \to &
{\rm e}^{-k}
\\[3mm]
\left(1 - {k \over n}\right)^{k}
& = &
{\rm e}^{k\ln\left(1\ -\ k/n\right)}
=
{\rm e}^{k\left(-\ k/n\ -\ k^{2}/2n^{2} - \cdots\right)}
& \to &
1
\end{array}
$$
A: for the first asks
$$\lim_{n \to \infty}{\left(1-\frac{k}{n}\right)}^n=e^{\lim_{n \to \infty}n\ln{\left(1-\frac{k}{n}\right)}}.$$ Now, making the change of variables $\frac{1}{n}=t,$ then 
$$e^{\lim_{n \to \infty}n\ln{\left(1-\frac{k}{n}\right)}}=e^{\lim_{t \to 0}\left[\frac{1}{t}\ln{\left(1-kt\right)}\right]}.$$
We have a indeterminate from of type $0/0$. Now, we can apply the L'Hospital's rule and gain 
$$e^{\lim_{t \to 0}\frac{\ln{\left(1-kt\right)}}{t}}=e^{\lim_{t\to 0}\frac{\frac{d}{dt}\ln(1-kt)}{\frac{d}{dt}t}}=e^{\lim_{t\to 0}\frac{1}{1-kt}(-k)} =e^{-k}.$$
We can use the same approach to solve the second issue..
A: It is a fact that $$\lim_{x \to \infty}{\left(1+\frac{1}{x}\right)}^x=e$$
You can find a proof of that here: http://aleph0.clarku.edu/~djoyce/ma122/elimit.pdf
Now, $$\lim_{n\to\infty}{\left(1-\frac{k}{n}\right)}^n=\lim_{n\to\infty}{\left[1+\left(-\frac{k}{n}\right)\right]}^n$$
$$=\left[\lim_{n\to\infty}{\left[1+\left(-\frac{k}{n}\right)\right]}^{(-{n}/{k})}\right]^{-k}=e^{-k}$$
For the second limit, k is just a number, it does not tend to infinity, so the previous formula doesnt apply anymore.
So, because $\frac{1}{n}\to 0$, we have
$$\lim_{n\to\infty}\left(1-\frac{k}{n}\right)^k=1^k=1$$
