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 ?
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 &amp;=\sum_{k&gt;n}\frac{x^k}{k!}+\sum_{k=0}^n\left(\frac1{k!}-\binom nk\frac1{k!}\right)x^k \\&amp;=\sum_{k=0}^\infty a_{n,k}\frac1{k!}x^k,\end{align}where$$a_{n,k} = \begin{cases} 1-\left(1-\frac1n\right) \cdots\left(1-\frac{k-1}n\right)&amp; \text{ if }k \leqslant n\\1&amp;\text{ otherwise.}\end{cases}$$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 $\varepsilon&gt;0$ and choose $N\in\mathbb N$ such that, for $n\geqslant N$, $\sum_{k=n+1}^\infty\frac1{k!}b^k&lt;\frac\varepsilon2$. Then we have\begin{align}
\left\lvert e^x-\left(1+\frac xn\right)^n\right\rvert&amp;\leqslant\left\lvert\sum_{k=0}^N a_{n,k}\frac1{k!}x^k\right\rvert+\frac\varepsilon2 \\&amp;\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$$(\forall x\in[a,b]):\left\lvert e^x-\left(1+\frac xn\right)^n\right\rvert\leqslant\varepsilon.$$
Hint $:$ Observe that $f(x) = \lim\limits_{n \rightarrow \infty} \left (1 + \frac x n \right )^n = e^x,$ for all $x \in \Bbb R.$