Expanding a function with $\frac{\sin(nx)}{x}$, and then recovering the delta function behaviour (Quick note; I'm a physicist so it means I might have crowbarred some maths into an equation that shouldn't be there, resulting in the problems I'm having.)
If we have
\begin{equation}
I= \int^{\infty}_{-\infty} \lim_{n \rightarrow \infty}\left(  \frac{\sin (n x)}{x} \right) f(x) \mathrm dx=\lim_{n\rightarrow \infty}I_{n},
\end{equation}
then through a change of variables, $nx=y$
\begin{equation}
I_{n} = \int^{\infty}_{-\infty} \left(  \frac{\sin (n x)}{x} \right) f(x) \mathrm dx =  \int^{\infty}_{-\infty} \left(  \frac{\sin (y)}{y} \right) f \left(\frac{y}{n} \right) \mathrm dy,
\end{equation}
and so
\begin{equation}
I = \int^{\infty}_{-\infty}\left(  \frac{\sin (y)}{y} \right) f \left( 0 \right) \mathrm dy = \pi f \left( 0 \right) .
\end{equation}
This replicates the Dirac delta function, meaning that at this limit we can say
\begin{equation}
\lim_{n \rightarrow \infty} \left( \frac{\sin (n x)}{x} \right) \rightarrow \pi \delta(x).
\end{equation}
I would like to expand the test function, and then, through taking the limit, recover the delta function property of our function. 
\begin{equation}
I_{n} = \int^{\infty}_{-\infty} \left(  \frac{\sin (n x)}{x} \right) f(x) \mathrm dx =  \int^{\infty}_{-\infty} \left(  \frac{\sin (y)}{y} \right) f \left(\frac{y}{n} \right) \mathrm dy,
\end{equation}
Putting the function through a Taylor expansion,
\begin{equation}
\int^{\infty}_{-\infty} \left(  \frac{\sin (y)}{y} \right) f \left(\frac{y}{n} \right) \mathrm dy= \int^{\infty}_{-\infty} \left(  \frac{\sin (y)}{y} \right) \left( f \left(0\right) + \frac{y}{n}f^{\prime} \left(0\right) + \frac{y^{2}}{n^{2}}f^{\prime \prime} \left( 0 \right)  + \mathcal O\left( \frac{y^{3}}{n^{3}} \right) \right) \mathrm dy,
\end{equation}
This integral clearly diverges, I could take the limit at this stage, but I don't understand why this expansion would not work. 
Ultimately, I would like a expression that is the result of an integral with the $\frac{\sin(nx)}{x}$ function, with extra terms from the expansion. This expression would then reconstruct the delta function behaviour as $n \rightarrow \infty$, with the extra terms decaying to zero. I understand this might not be possible, and if so I'd like to know why!
 A: Your series has non meaning as the limits of the sin and cos functions, for the argument running to infinity, do not exist. Things are quite different if this is seen in the sense of distributions. Firstly, we note the obvious fact that
$$
   \int_{-\infty}^\infty\frac{\sin kx}{x}dx=
\int_{-\infty}^\infty\frac{\sin x}{x}dx=\pi.
$$
Then, you can just Taylor expand the test function and write
$$
   \int_{-\infty}^\infty\frac{\sin kx}{x}f(x)dx=
\pi f(0)+\int_{-\infty}^\infty\frac{\sin kx}{x}xf'(0)dx
+\frac{1}{2}\int_{-\infty}^\infty\frac{\sin kx}{x}x^2f''(0)dx+\ldots.
$$
That integrals can only be seen as distributions but
$$
  \int_{-\infty}^\infty\sin kxdx=-i\pi[\delta(k)-\delta(-k)]=0
$$
as the Dirac distribution is even. Then,
$$
  \int_{-\infty}^\infty x\sin kxdx=\frac{1}{2i}\int_{-\infty}^\infty
x[e^{ikx}-e^{-ikx}]=-2\pi\frac{d}{dk}\delta(k)
$$
and
$$
  \int_{-\infty}^\infty x^2\sin kxdx=\frac{1}{2i}\int_{-\infty}^\infty
x^2[e^{ikx}-e^{-ikx}]=i\pi\frac{d^2}{dk^2}[\delta(k)-\delta(-k)]=0
$$
and so on, noting that only even terms of the Taylor expansion survive as distributions. In the end you will get a distribution series in $k$ with all the terms, with the exception of the leading one, that will go to zero when $k$ runs to infinity. This limit can only be attached a meaning in the sense of distributions.
