A doubt about $\lim_{y\to \infty} \frac{\sin (yx)}{x}= \delta(x)$

Question

I understand $$\lim_{x\to 0} \frac{\sin (yx)}{x}= y$$ Next, $$\lim_{y\to \infty}\left(\lim_{x\to 0} \frac{\sin (yx)}{x}\right)=\infty=\delta(0)~~~~(1)$$ which is very convincing. But then I should get $$\lim_{y\to \infty}\left(\lim_{x\to a} \frac{\sin (yx)}{x}\right)=0=\delta(a),~~~~(2)$$ or $x=a \ne 0$. But I don't get it. Please help me to get (2) in a simple way.

Answer 1

What you have doesn't make any sense. The definition of Dirac delta is that it is a distribution defined by

$$\delta_a(f) = f(a)$$

We can interpret the limit, however, by checking

$$\lim_{y\to\infty}\int_{-\infty}^\infty \frac{\sin yx}{x}f(x)\:dx$$

for suitable $f$ (Take $f$ to be Schwartz). Use the substitution $u = yx$ to get

$$\lim_{y\to\infty}\int_{-\infty}^\infty \frac{\sin u}{u}f\left(\frac{u}{y}\right)\:du = f(0) \int_{-\infty}^\infty\frac{\sin u}{u}du = \pi f(0)$$

by Dominated Convergence. So we can reasonably interpret the limit as being

$$\lim_{y\to\infty}\frac{\sin yx}{x} = \pi\delta(x)$$