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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 ?

Todd
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    Actually, I don't think that $f_n\to e^x$ uniformly on compact sets. A possible way to solve this problem, is to prove that $f(x)=\sum_{k=0}^\infty \frac{x^k}{k!}$ for all $x\in\mathbb R$. This will even prove that $f\in \mathcal C^\infty $. An other way can be found here – Surb Dec 01 '19 at 10:06

4 Answers4

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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$$(\forall x\in[a,b]):\left\lvert e^x-\left(1+\frac xn\right)^n\right\rvert\leqslant\varepsilon.$$

José Carlos Santos
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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.

Uday Khanna
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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.

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

math maniac.
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