Uniform Convergence to the Exponential Function over a Compact Interval I'm trying to show that the sequence of functions $f_n(x)=(1+(x/n))^n$ converges uniformly to $f(x)=e^x$ over any compact interval of the real line. We're assuming that it converges pointwise. Here is what I have so far:
Assume, for contradiction, that $f_n$ does not converge uniformly to $f$ over a compact interval $I$. Using the negation of the definition of uniform convergence, this means that there is an $\epsilon>0$ such that for each natural number $n$, there is an $x_n$ and an ${f_m}_n$ such that $m_n\ge n$ and $|{f_m}_n(x_n)-f(x_n)|>\epsilon$.
Assume there is a countable number of distinct $x_n$'s because if there were only a finite number of them, then, by pointwise convergence, we could take $M=\max(N_{\epsilon,x_i} : n\ge N_{\epsilon,x_i}\Rightarrow|f_n(x_i)-f(x_i)|\le\epsilon)$ and $m\ge M$ would give us uniform convergence.
By Bolzano-Weierstrass, this sequence of distinct $x_n$'s converges to a point $x$ in $I$, since $I$ compact. By triangle inequality, $$|{f_m}_n(x_n)-f(x_n)|\le |{f_m}_n(x_n)-f(x)|+|f(x)-f(x_n)|$$
Since $f$ is continuous everywhere, in particular, at $x$, and since, given $\delta$, there is an integer $N$ such that $n\ge N\Rightarrow |x_n-x|\le \delta$, we have $|f(x)-f(x_n)|\le \epsilon/2$ for an infinite number of $x_n$'s.
I run into trouble when I try to show that there is an $x_n$ such that $|{f_m}_n(x_n)-f(x)|\le \epsilon/2$. Intuitively, this should be true as $n$ gets very large, since we have pointwise convergence, continuity, and $x_n$'s converging to $x$. However, I am having a lot of trouble showing this rigorously. Can anybody give me some help here, or tell me that I'm on a wrong path? 
The end goal is the contradiction of there being an $x_j$ such that $|{f_m}_j(x_j)-f(x_j)|>\epsilon$ and $|{f_m}_j(x_j)-f(x_j)|\le \epsilon$. Any help at all would be much appreciated! 
 A: Here is a direct proof for another perspective.  
Start with the case $x \in [0,b]$.
Using the inequality $\ln(1+y) \leqslant y$, we have for $0 \leqslant y < 1$,
$$1+y \leqslant e^y = \sum_{k=0}^{\infty} \frac{y^k}{k!} <  \sum_{k=0}^{\infty} y^k = \frac1{1-y},$$
Take $y = x/n$. It follows that for $n$ sufficiently large
$$1 + \frac{x}{n} \leqslant e^{x/n} < \left(1 - \frac{x}{n}\right)^{-1},$$
and
$$\left(1 + \frac{x}{n}\right)^n \leqslant e^x < \left(1 - \frac{x}{n}\right)^{-n}.$$
Hence,
$$0 \leqslant e^x - \left(1+ \frac{x}{n}\right)^n= e^x\left[1 - \left(1+ \frac{x}{n}\right)^ne^{-x}\right]\\ \leqslant e^x\left[1 - \left(1+ \frac{x}{n}\right)^n\left(1- \frac{x}{n}\right)^n\right]\\ = e^x\left[1 - \left(1- \frac{x^2}{n^2}\right)^n\right].$$
Using Bernoulli's inequality,
$$\left(1- \frac{x^2}{n^2}\right)^n \geqslant 1-n\frac{x^2}{n^2}= 1-\frac{x^2}{n},$$
and it follows that
$$0 \leqslant e^x - \left(1+ \frac{x}{n}\right)^n \leqslant e^x \frac{x^2}{n} \leqslant e^b \frac{b^2}{n}.$$
Since 
$$\lim_{n \to \infty}e^b \frac{b^2}{n} = 0,$$
it follows that $\displaystyle \left(1+ \frac{x}{n}\right)^n$ converges uniformly to $e^x$ on $[0,b]$.
Using a similar argument, you can prove uniform convergence on an interval $[-a,0]$ and combining the results show uniform convergence on any compact interval in $\mathbb{R}$.
