How is the following limit is true?

Question

$ {(1)\space \displaystyle \lim _{a\to 0}{\frac {\sin \left({\frac {\pi x}{a}}\right)}{\pi x}}=\lim _{a\to 0}{\frac {1}{a}}\operatorname {sinc} \left({\frac {x}{a}}\right)=\delta (x).} $ How is that true, giving that the delta is defined that way $(2)\space \delta (x)={\begin{cases}+\infty ,&x=0\\0,&x\neq 0\end{cases}}.$ And why Fourier Transform in our course we use the definition number (2) even though the definition (1) align better with Fourier Transform of cosine and sine.

Answer 1

The $\delta$ distribution is not defined the way you write. A more correct definition is $$\int_{-\infty}^{\infty} \delta(x) \, f(x) \, dx = f(0),$$ where $f$ is a "nice" function.

To show that $\frac{\sin \frac{\pi x}{a}}{\pi x} \to \delta(x)$ we need to show that $$ \int_{-\infty}^{\infty} \frac{\sin \frac{\pi x}{a}}{\pi x} \, f(x) \, dx \to f(0) $$ when $a\to 0$ and $f$ is a "nice" function.

And, using the substitution $y=\frac{\pi x}{a},$ we have $$ \int_{-\infty}^{\infty} \frac{\sin \frac{\pi x}{a}}{\pi x} \, f(x) \, dx = \int_{-\infty}^{\infty} \frac{\sin y}{ay} \, f(\frac{ay}{\pi}) \, \frac{a\,dy}{\pi} = \frac{1}{\pi} \int_{-\infty}^{\infty} \frac{\sin y}{y} \, f(\frac{ay}{\pi}) \, dy \to \frac{1}{\pi} \int_{-\infty}^{\infty} \frac{\sin y}{y} \, f(0) \, dy \\ = \frac{1}{\pi} \int_{-\infty}^{\infty} \frac{\sin y}{y} \, dy \, f(0) = f(0) $$ since $\int_{-\infty}^{\infty} \frac{\sin y}{y}=\pi.$