Proofs that $f^{(100)}(0)= -\frac1{101}$, where $f(x) = \frac{\sin x}x$ and $f(0)=1$ 
4, Let $$f(x)=\begin{cases}
\dfrac{\sin{x}}{x}&x\neq 0\\
1&x=0
\end{cases}$$
  Find the value $f^{(100)}(0)$.

I find that $$f^{(100)}(0)=-\dfrac{1}{101}$$
by noticing
$$\dfrac{\sin{x}}{x}=\sum_{k=1}^{\infty}(-1)^{2k-1}\dfrac{1}{(2k-1)!}x^{2k-2}$$
Do you know other nice methods?
 A: Here's a reasonably short alternative; it uses a bit of complex number theory.
Note that by Euler's formula, $\dfrac{\sin x}x = \operatorname{im}\left(\dfrac{e^{ix}}{x}\right)$.
Now we introduce a new variable $k$ (with the intention that $k = 1$), and note that:
$$\frac{e^{ikx}}x =\int_0^k -ie^{i\kappa x}\,\mathrm d\kappa$$
Interchanging differentiation, integration and the imaginary part $\operatorname{im}$, we obtain that:
$$\begin{align}
f^{(100)}(0) &= \left.\frac{\mathrm d^{100}}{\mathrm dx^{100}} \left(\operatorname{im}\int_0^k -ie^{i\kappa x}\,\mathrm d\kappa\right) \right\vert_{x=0,k=1}\\
&=\left.\operatorname{im}\int_0^k \frac{\mathrm d^{100}}{\mathrm dx^{100}} \left(-ie^{i\kappa x}\right)\,\mathrm d\kappa\right\vert_{x=0,k=1}\\
&= \left.\operatorname{im} \int_0^k-i(i\kappa)^{100}e^{i\kappa x}\,\mathrm d\kappa\right\vert_{x=0,k=1}\\
&= \left.\operatorname{im}\int_0^k -i\kappa^{100}\,\mathrm d\kappa\right\vert_{k=1}\\
&= \left.\operatorname{im} \frac{-ik^{101}}{101}\right\vert_{k=1}\\
&= -\frac1{101}
\end{align}$$
where it was used that $i^{100} = 1$.
Note how crucial it was that we knew we were evaluating at $x = 0$; for unknown $x$, the integral can't be resolved in elementary terms.
