Swapping limits: $\lim_{h\to 0}\lim_{n\to \infty}\frac {(1+1/n)^{hn}-1}{h}=\lim_{n\to\infty}\lim_{h\to 0}\frac {(1+1/n)^{hn}-1}{h}$ Almost a year ago I asked the question: How to differentiate $e^x$?
And in the accepted answer, the following equality appeared:
$$\lim_{h\to 0}\lim_{n\to \infty}\frac {(1+1/n)^{hn}-1}{h}=\lim_{n\to\infty}\lim_{h\to 0}\frac {(1+1/n)^{hn}-1}{h}$$
What allows one to switch limits in this case?
 A: Uniform convergence of the inner limit is sufficient to justify the switch.
You can show using $(1 + 1/n)^{n} < e < (1-1/n)^{-n}$ that for $h \in [-\delta, \delta]$ 
$$\left|\frac{e^h - \left(1 + 1/n\right)^{nh}}{h}\right| < \frac{e^{|h|}}{n} < \frac{e^\delta}{n}.$$
For $h > 0$, using the above inequalities and the Bernoulli inequality
$$0 < e^h - (1+1/n)^{nh} = e^h[1 - e^{-h}(1+1/n)^{nh}] \\ \leqslant e^h[1 - (1 -1/n^2)^{nh}] \\ \leqslant e^h \frac{nh}{n^2} \\ \frac{he^h}{n}.$$
(Make a similar estimate for $h < 0$.)
Hence, we have 
$$\lim_{n \to \infty}\frac{(1+1/n)^{nh}-1}{h} = \frac{e^h-1}{h},$$
uniformly for $h \in [-\delta,\delta]$.
A: For a sequence of continuous functions $f_n(x),$ $$\lim_{x \rightarrow x_0} \lim_{n \rightarrow \infty} f_n(x) = \lim_{n \rightarrow \infty} \lim_{x \rightarrow x_0} f_n(x)$$ is the same as saying that the pointwise limit of $f_n$ is continuous. The usual criterion is for $(f_n)_n$ to converge uniformly on compact intervals.
Once you have agreed that $\lim_{h \rightarrow 0} \frac{(1 + 1/n)^{nh} - 1}{h}$ exists for all $n$, and that the results are bounded in $n$, you can conclude that the sequence is uniformly bounded and equicontinuous and you can apply Arzela-Ascoli's theorem.
