Use the Binomial Formula

\[(a+b)^{n}=\sum_{k=0}^{n}\left(\begin{array}{l}n \\k\end{array}ight) a^{k} b^{n-k}\]

to show

(a)

\[e=\lim _{n ightarrow \infty}\left(1+\frac{1}{n}ight)^{n}=\sum_{k=0}^{\infty} \frac{1}{k !}\]

Hint:

\[\begin{aligned}\left(1+\frac{1}{n}ight)^{n} & =\sum_{k=0}^{n} \frac{n !}{k !(n-k) !} \frac{1}{n^{k}} \\& \ldots \\& =\sum_{k=0}^{n} \frac{1}{k !} B(k, n)\end{aligned}\]

where

\[B(k, n)=\frac{n(n-1) \ldots(n-(k-1))}{n^{k}} \text { for each } k\]

(b) Use the Binomial Formula to show

\[e^{x}=\lim _{n ightarrow \infty}\left(1+\frac{x}{n}ight)^{n}=\sum_{k=0}^{\infty} \frac{x^{k}}{k !}\]

Hint: Let \(n=m x\), and evaluate \((1+x / n)^{n}=(1+x /(m x))^{m x}\).