If $f(x)=\lim_{n\to \infty }\left(1+\frac{x}{n}\right)^n$ prove that $f$ is continuous. I'm stuck on this problem. I proved that $f_n(x)=\left(1+\frac{x}{n}\right)^n$ converge for all $x\in\mathbb R$. Set $$f(x)=\lim_{n\to \infty }f_n(x).$$ I'm now trying to prove that $f$ is continuous. I tried to prove that $f_n\to f$ uniformly on compact but I failed. I tired as follow :
Set$$g_n(x)=f_n(x)-e^x.$$
Then $$g_n'(x)=\left(1+\frac{x}{n}\right)^{n-1}-e^x.$$
Now, how can I solve $g_n'(x)=0$ to find it's maximum ? Also, I tried to show that $g_n\to 0$ uniformly on all compact, but also, I failed. Any idea ?
 A: Yes, that sequence converges uniformly on every compact subset of $\mathbb R$. That is sufficient to prove that the limit is continuous. Of course, it is enough to prove that the convergence is uniform on any interval $[a,b]$.
Note that\begin{align}e^x-\left(1+\frac xn\right)^n &=\sum_{k>n}\frac{x^k}{k!}+\sum_{k=0}^n\left(\frac1{k!}-\binom nk\frac1{k!}\right)x^k \\&=\sum_{k=0}^\infty a_{n,k}\frac1{k!}x^k,\end{align}where$$a_{n,k} = \begin{cases} 1-\left(1-\frac1n\right) \cdots\left(1-\frac{k-1}n\right)& \text{ if }k \leqslant n\\1&\text{ otherwise.}\end{cases}$$Note also that $0 \leqslant a_{n,k} \leqslant 1$ for all $k,n\in\mathbb N$ and, for any fixed $k$, $\lim_n a_{n,k}=0$.
Let $\varepsilon>0$ and choose $N\in\mathbb N$ such that, for $n\geqslant N$, $\sum_{k=n+1}^\infty\frac1{k!}b^k<\frac\varepsilon2$. Then we have\begin{align}
\left\lvert e^x-\left(1+\frac xn\right)^n\right\rvert&\leqslant\left\lvert\sum_{k=0}^N a_{n,k}\frac1{k!}x^k\right\rvert+\frac\varepsilon2 \\&\leqslant\sum_{k=0}^N a_{n,k}\max\left\{1,b^N\right\}+\frac\varepsilon2\end{align}
Now choose $N^\ast\geqslant N$ such that $\sum_{k=0}^N a_{n,k}\leqslant \frac\varepsilon{2\max\left\{1,b^N\right\}}$ for every $n\geqslant N^\ast$, in order to get$$(\forall x\in[a,b]):\left\lvert e^x-\left(1+\frac xn\right)^n\right\rvert\leqslant\varepsilon.$$
A: Instead of a limit of sequence of functions we can also consider a sequence in the domain.
Let $x_{0}\in\mathbb{R}$ and $(x_{k})\rightarrow x_{0} $ in $\mathbb{R}$
Then  $ \lim_{k\rightarrow\infty}(\lim_{n\rightarrow\infty}\left(1+\frac{x_{k}}{n}\right)^n)$ =  $\lim_{n\rightarrow\infty}(\lim_{k\rightarrow\infty}\left(1+\frac{x_{k}}{n}\right)^n)$ = $\lim_{n\rightarrow\infty}\left(1+\frac{x_{0}}{n}\right)^n)$ as $n\geq1$
Which shows that f is continuous by sequential criteria of continuity.
A: 
Hint Prove that $f_n(x)$ are increasing with respect to $n$ (after some $n_0$) and then use the fact that monotonicity implies uniform convergence on compact sets.

A: Hint $:$ Observe that $f(x) = \lim\limits_{n \rightarrow \infty} \left (1 + \frac x n \right )^n = e^x,$ for all $x \in \Bbb R.$
