Show that $$f_n(x) = \frac{\sin nx}{\pi x}$$ converges to $\delta$ in the sense of distribution. I know the same question have been asked here. But I found some hint from Arfken Mathematical Methods for physicists which is
Show that $$\lim_{n\to\infty}\int_{-\infty}^{\infty}f(x)\frac{\sin nx}{\pi x}dx=0$$ Assume that $f(x)$ is continuous at $x = 0$ and vanishes as $x → ±∞$.
Hint. Replace $x$ by $y/n$ and take $\lim\ n → ∞$ before integrating.
with this hint I found the following solution(simple to grasp)
from this site. Everything in the above solution is a pretty much convincing to me except while substitution the denominator should be $\pi y/n$. But having that "$n$" the limit become "ugly" and the result would be different. So, how can we show $\lim_{n\to\infty}\int_{-\infty}^{\infty}f(x)\frac{\sin nx}{\pi x}dx=0$ using the hint.
Let $y = nx$, then $x = y/n$ and $dx = dy/n$, hence $$\int_{-\infty}^{\infty}f(x)\frac{\sin nx}{\pi x}dx=\int_{-\infty}^{\infty}f(y/n)\frac{\sin y}{\pi (y/n)}d(y/n)=\int_{-\infty}^{\infty}f(y/n)\frac{\sin y}{\pi y}dy,$$ which allows you to pass to the limit.
Does this answer your question?