Help with dirac delta derivative Prove that
$\sigma_{n}=n^{3}$sin$[nx+\frac{1}{n}]\delta(nx)-n\delta(x-\frac{1}{n})$
has the distributional limit 
$\lim_{n\to\infty } \sigma_{n}=\delta'(x).$
I know that $\delta'(x)=\lim_{h\to0}\frac{\delta(x+h)-\delta(x)}{h}$ but i fail to see how I can use this. Any hints, guidence or help is welcomed.
 A: Given the two "facts"
$$ \delta(c x) = \frac{1}{|c|} \delta(x), \text{ and } f(x) \delta(x) = f(0) \delta(x)$$
for $c$ constant and $f$ continuous, we obtain
$$\lim_{n\to\infty}\sigma_n = \lim_{n\to\infty} n^2 \sin(n^{-1}) \delta(x) - n \delta(x-1/n).$$
Now set $h=n^{-1}$ and observe that $h\to0$ when $n\to\infty$, so
$$\lim_{n\to\infty}\sigma_n = \lim_{h\to0} \frac{\overbrace{h^{-1}\sin h}^{1+O(h^2)}\, \delta(x) - \delta(x-h)}{h}=\lim_{h\to0} \frac{\delta(x) - \delta(x-h)}{h} + O(h)  = \delta'(x)  $$
A: Starting from the two following properties of $\delta$
$$\begin{align}\delta(cx)&={1\over |c|}\delta(x)\\f(x)\delta(x)&=f(0)\delta(x)\end{align}$$
we can write
$$E(x,n)=n^3\sin\left(nx+{1\over n}\right)\delta(nx)-n\delta\left(x-{1\over n}\right)=n\sin{1\over n}\times n\delta(x)-n\delta\left(x-{1\over n}\right)$$
Now consider that as $n\to\infty$ we have $n\sin{1/n}\to 1$ and so
$$\begin{align}\lim_{n\to\infty}E(x,n)&=\lim_{n\to\infty}n\left(\delta(x)-\delta\left(x-{1\over n}\right)\right)\\&=\lim_{h\to 0}{\delta(x)-\delta\left(x-h\right)\over h}\end{align}$$
