How to calculate $\lim \limits_{n \to \infty}(1 +\frac{a}{n})^n$? I know what solution is $e^a$ but I don't know how to calculate this limit:
$$\lim \limits_{n \to \infty}\left(1 + \frac{a}{n}\right)^n=e^a$$
Can someone explain the steps to me?
 A: $$\lim \limits_{x \to +\infty} (1+a/x)^x = e^{\lim \limits_{x \to +\infty} x \log (1+a/x)}$$
Then apply l'Hôpital's rule to $\frac{\log (1+a/x)}{1/x}$ to get $$\frac{\frac{1}{1+a/x} \cdot \frac{-a}{x^2}}{\frac{-1}{x^2}} = \frac{a}{1+a/x} = a$$
A: Use $\lim_{n\rightarrow\infty} (1+1/n)^n=e$.
Let $m=n/a$ then $\lim_{n\rightarrow\infty} (1+a/n)^n=\lim_{m\rightarrow \infty} (1+1/m)^{ma}=e^a$. 
A: Let $y=1/n\implies \lim \limits_{n \to \infty}\left(1 + \frac{a}{n}\right)^n=\lim_{y\to 0^+}(1+ay)^{\frac{1}{y}}$
Let $$A=\lim_{y\to 0^+}(1+ay)^{\frac{1}{y}}$$ $$\implies \log A=\lim_{y\to 0^+}\frac{\log(1+ay)}{y}=\lim_{y\to 0^+} \frac{a}{(1+ay)}=a$$ (applying L'Hopital's Rule) $$\implies \log A=a\implies A= e^a$$
Thus, $$\lim \limits_{n \to \infty}\left(1 + \frac{a}{n}\right)^n=e^a$$
A: Here is a more tedious way:
Let $c_{n,k} = \begin{cases}
0 && k>n \\ \binom{n}{k} \frac{1}{n^k} && \text{otherwise} \end{cases}$
Note that $0 \leq c_{n,k} \leq \frac{1}{k!}$, $c_{n+1,k} \geq c_{n,k}$, and $\lim_n c_{n,k} =\frac{1}{k!}$. (Showing these facts is the tedious part.) And, of course, $(1+\frac{a}{n})^n = \sum_{k=0}^\infty c_{n,k} a^k$.
Now note that $|e^a - (1+\frac{a}{n})^n| \leq \sum_{k=0}^\infty |\frac{1}{k!}-c_{n,k}| |a^k| \leq \sum_{k=0}^\infty \frac{1}{k!} |a^k| \leq e^{|a|}$. (The latter uses the fact that $0 \leq \frac{1}{k!}-c_{n,k} \leq \frac{1}{k!}$.)
Let $\epsilon>0$ and choose $N$ such that $\sum_{k>N} \frac{1}{k!} |a^k| < \frac{\epsilon}{2}$. Let $M = \max(1, |a|,...,|a|^N)$. Now choose $N' \geq N$ such that $\max_{k=0,...,N} |\frac{1}{k!}-c_{n,k}| < \frac{1}{M(N+1)} \frac{\epsilon}{2}$ whenever $n>N'$.
Then we have $|e^a - (1+\frac{a}{n})^n| \leq \sum_{k=0}^\infty |\frac{1}{k!}-c_{n,k}| |a^k| \leq \sum_{k=0}^N |\frac{1}{k!}-c_{n,k}| |a^k|+ \sum_{k>N} |\frac{1}{k!}-c_{n,k}| |a^k|$. Both terms on the right hand side are bounded by $\frac{\epsilon}{2} $, hence we have $|e^a - (1+\frac{a}{n})^n| < \epsilon$, for $n> N'$.
