Using the Dirac delta function for taking limits

Question

In trying to calculate:

$$\displaystyle \lim\limits_{a\to0} \frac{2x}{a^3} e^{-\frac{x^2}{a^2}}. \tag{1}$$

I initially tried to make an argument with the taylor series (looking at the limit for each term individually), but there were some edge cases that were difficult to deal with( when x=0) and the whole argument was somewhat ugly. I've been advised some properties of the dirac-delta function (and it's derivative) can bee applied to this problem, but I don't quite see how. The function itself does not behave like a dirac-delta, nor is any convenient piece of it (so far as I can see).

How would using the function be relevant to these types of problems?

Answer 1

Well, there is a so-called heat kernel representation of a nascent delta function $$\delta(x)~=~\lim_{a\to 0^+} \delta_a(x) , \qquad \delta_a(x)~:=~\frac{1}{\sqrt{\pi}a}e^{-\frac{x^2}{a^2}}, $$ $$\delta^{\prime}(x)~=~\lim_{a\to 0^+} \delta^{\prime}_a(x) , \qquad \delta^{\prime}_a(x)~=~-\frac{2x}{\sqrt{\pi}a^3}e^{-\frac{x^2}{a^2}}. $$

In other words. OP's formula (1) is a representation of the derivative of the Dirac delta distribution (times a factor minus square root of pi).

Answer 2

One usually uses this limit to show properties of an approximation to the Dirac Delta. However, here is a hint to show this limit.

Hint: Let $u=\frac1a$, then $$ \begin{align} \left|\frac{2x}{a^3}e^{-\frac{x^2}{a^2}}\right| &=\frac{2\left|xu^3\right|}{e^{x^2u^2}}\\ &\le\frac{2\left|xu^3\right|}{1+x^2u^2+\frac{x^4u^4}2}\\ \end{align} $$

Answer 3

Using Dirac's delta function here would be an overkill. From what you wrote, we see that $a$ is the variable of this limit and $x$ is a parameter. Simply consider two cases for the parameter $x$:

• If $x=0$, then $\displaystyle \lim\limits_{a\to0} \frac{2x}{a^3} e^{-\frac{x^2}{a^2}}=\displaystyle \lim\limits_{a\to0} 0=0$.

• If $x\neq0$, then $\displaystyle \lim\limits_{a\to0} \frac{2x}{a^3} e^{-\frac{x^2}{a^2}}=2x\lim\limits_{a\to0} \frac{e^{-\frac{x^2}{a^2}}}{a^3}$, and then complete it using L'Hôpital's rule or Taylor series, or just recall than exponential functions grow or decay much faster than power functions.

Answer 4

$f(x) = 2x e^{-x^2}$ then you are looking at $\lim_{n \to \infty} f_n$ where $f_n(x)=n^{2} f(nx)$, which converges locally uniformly to $0$ for $|x| > \epsilon$ but not around $x=0$.

Note $n^2 f(nx ) = g_n^{''}(x)$ where $g_n(x) = g(nx)$ and $g(x) = \int_0^x y f(y)dy, f = g''$. It is not hard to see $\lim_{n \to \infty} g(nx) = 1_{x > 0} g(+\infty)+1_{x < 0} g(-\infty)$ where the convergence is locally uniform away from $x= 0$ and is bounded and thus $L^1_{loc}$ around $x=0$.

Therefore $\lim_{n \to \infty} g_n$ converges to $1_{x > 0} g(+\infty)+1_{x < 0} g(-\infty)$ in the sense of distributions,

$\lim_{n \to \infty} g_n'$ converges in the sense of distributions to its distributional derivative $(g(+\infty)-g(-\infty)) \delta$,

and $$\lim_{n \to \infty} f_n=\lim_{n \to \infty}g_n{''}=(g(+\infty)-g(-\infty)) \delta^{'}$$ in the sense of distributions.