Double limit of the nth derivative of $f(x)=\exp(\sqrt{x})+\exp(-\sqrt{x})$ I have tried to compute $$\lim\limits_{n\to \infty}\lim\limits_{x\searrow 0}f^{(n)}(x)$$ for $f:[0,\infty)\to \mathbb{R},~f(x)=\exp(\sqrt{x})+\exp(-\sqrt{x})$.
By noticing that $f$ and its derivatives satisfy $2f'(x)+4xf''(x)=f(x)$ and differentiating successively in this equality one can deduce a recurrence relation for $\lim\limits_{x\searrow 0}f^{(n)}(x)$ and thus obtain $\lim\limits_{x\searrow 0}f^{(n)}(x)=2\cdot\frac{n!}{(2n)!}$, but only assuming that the limit $\lim\limits_{x\searrow 0}~xf^{(n)}(x)$ is $0$ for any $n\in \mathbb{N}$, which I could not prove rigorously, but I "feel" that is true.
Of course, any other idea for the computation of the above double limit is welcomed.
 A: Note that for $f(x)= e^{\sqrt{x}}+e^{-\sqrt{x}}$, we can write
$$f(x)=2\sum_{k=0}^\infty \frac{x^{k}}{(2k)!}\tag1$$
Differentiating the power series on the right-hand side of $(1)$ $n$ times yields
$$f^{(n)}(x) = 2\sum_{k=n}^\infty \frac{k(k-1)\cdots (k-n+1)}{(2k)!}x^{k-n} \tag2$$
Letting $x\to 0^+$, we find that
$$\lim_{x\to 0^+}f^{(n)}(x)=2\frac{n!}{(2n)!}$$
Finally, letting $n\to \infty$ yields the coveted result
$$\lim_{n\to \infty}\lim_{x\to 0^+}f^{(n)}(x)=0$$
A: $\exp x$ has a series development, valid for all $x\in\Bbb C$: $$\exp x=\sum_{n=0}^\infty \frac{x^n}{n!}$$
Therefore, for $x\ge0$,
$$f(x)=\exp(\sqrt x)+\exp(-\sqrt x)=\sum_{n=0}^{\infty}\frac{x^{n/2}(1+(-1)^n)}{n!}=2\sum_{n=0}^\infty \frac{x^n}{(2n)!}$$
This series also has infinite radius, and $f$ is entire, and therefore $C^{\infty}$.
We can find it's $k$-th derivative:
$$f^{(k)}(x)=2\sum_{n=k}^{\infty}\frac{n!}{(n-k)!}\frac{x^{n-k}}{(2n)!}$$
Hence
$$\lim_{x\to0}f^{(k)}(x)=f^{(k)}(0)=\frac{2k!}{(2k)!}$$
And since $f$ is $C^{\infty}$, its derivative is finite at $0$, and $\lim_{x\to0} xf^{(k)}(x)=0$.
