I saw the proofs on the derivative of $\frac{d e^x}{dx}=e^x$ from here and the one that was intriguing was this : $$e^x:=\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n \implies \frac{d(e^x)}{dx} = \lim_{n\to\infty} n\cdot \frac{1}{n}\left(1+\frac{x}{n}\right)^{n-1} = \lim_{n\to\infty}\left(1+\frac{x}{n}\right)^{n-1} = e^x$$
Now what I am curious is this:
What theorem allows you to take derivative while keeping the limit sign intact and take the limit later as above?
The sequence of derivatives converges locally uniformly. That allows the interchange of limit and differentiation.
If $(f_n)$ is a sequence of continuously differentiable functions that converges pointwise to $f$, and the sequence of derivatives $(f_n')$ converges locally uniformly to $g$, then $f$ is continuously differentiable and $f' = g$. For we have
$$f(y) - f(x) = \lim_{n\to\infty} f_n(y) - f_n(x) = \lim_{n\to\infty} \int_x^y f_n'(t)\,dt = \int_x^y g(t)\,dt.$$