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Suppose $$f(x)=e^{\sqrt{x}}+e^{-\sqrt{x}}$$ We know, from Taylor expansion that $$f(x)=2\lim_{n \to \infty}\left({1+\frac{x}{2!}+\frac{x^2}{4!}+\dots+\frac{x^n}{(2n)!}}\right)$$ for each $x \geq0$. What I want to know:

Question 1: Are we allowed to differentiate this equality with respect to $x$, to obtain the derivative of f(x)? That is, can we swap the derivative with the limit? $$f'(x)=2\lim_{n \to \infty}\left({\frac{1}{2!}+\frac{2x}{4!}+\dots+\frac{nx^{n-1}}{(2n)!}}\right)$$ for each $x \geq0$ ??

As I have read, some kind of uniform convergence is needed. How do i prove/ justify this uniform convergence?? Is it true always in these cases, for sequences of functions from Taylor expansions?

Question 2: If instead of the derivative, I want to compute the limit for $x \to 0 $ of $f(x)$ is the same true, I mean, can we swap the limits with respect to $n$ and $x$?

Is $$ \lim_{x \to 0+} \lim_{n \to \infty} f_{n}(x)=\lim_{n \to \infty} \lim_{x \to 0+} f_{n}(x)$$ ?

RobPratt
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amarius8312
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    Power series can be differentiated term by term, see for example https://math.stackexchange.com/q/484250/42969 – Martin R Mar 15 '25 at 16:57
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    John M.H. Olmsted - Advanced calculus-Prentice Hall (1961), page 184. – zkutch Mar 15 '25 at 17:05
  • For your other question, the general result for when two limits can be exchanged is the existence of the double limit. In your specific example that would be if there is some number $L$ such for every neighborhood $U$ of $L$, there exist $\delta > 0, N > 0$ such that for all $x \in (0,\delta), n > N$ it is true that $f_n(x) \in U$. When the double limit exists, then you can exchange the iterated limits. Otherwise, it might happen taht exchanging limits is okay, but in general it is not. – Paul Sinclair Mar 18 '25 at 18:44

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