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I see the comment by Henning Makholm may be relevant to my question on Derivative of sinc function. But I don't quite get the background. Could anybody show me to prove all the derivatives of $\mathrm{sinc}(x)$ are continuous at $x=0$ without omission of details? Thanks!

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I think Jyrki Lahtonen's comment in that thread would be more useful to you. – Neal Jul 6 '12 at 2:24

Expanding on @JyrkiLahtonen's comment:

Since $\sin x = \sum_{k=0}^{\infty} (-1)^k \frac{x^{2k+1}}{(2k+1)!}$, you have $\mathbb{sinc}\, x = \sum_{k=0}^{\infty} (-1)^k \frac{x^{2k}}{(2k+1)!}$. It is straightforward to show that the series converges uniformly on any bounded set, hence it is $C^{\infty}$.

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