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?

  • $\begingroup$ be careful, $$e=\lim_{n\to\infty }\left(1+\frac{x}{n}\right)^n$$ and not $$\lim_{n\to\infty }\left(1+\color{red}{\frac{n}{x}}\right)^n.$$ Moreover, $$\lim_{n\to\infty }\left(1+\frac{x}{n}\right)^{n-1}\neq e$$ with out more justification. Actually, it's correct, but you need to justify ! $\endgroup$ – idm Aug 30 '14 at 15:34
  • $\begingroup$ Woops! that's right. Thank you for pointing that out and making corrections. $\endgroup$ – Alby Aug 30 '14 at 15:39

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

$$f(y) - f(x) = \lim_{n\to\infty} f_n(y) - f_n(x) = \lim_{n\to\infty} \int_x^y f_n'(t)\,dt = \int_x^y g(t)\,dt.$$

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