Calculating $ \lim_{x\rightarrow\infty}\frac{\delta(0)}{x} $

Question

Does this limit go to $1$? I am not sure how to calculate it, because it contains the Dirac-delta generalized function on $ 0 $. $$ \lim_{x\rightarrow\infty}\frac{\delta(0)}{x} $$

I came across this limit by trying to "evaluate" the following expression: $$ \lim_{x\rightarrow\infty}\frac{\int_{-x/2}^{x/2} \delta(t)^2 dt}{x} $$

Answer 1

So the first step to solve a problem is to understand it. The Dirac delta is defined as a linear functional over continuous functions by $$ \langle \delta_0, \varphi\rangle = ∫\varphi(y)\,\delta_0(\mathrm{d}y) = \varphi(0) $$ In particular, one can multiply $\delta_0$ by a continuous function $f$ by the formula $$ \langle f\,\delta_0, \varphi\rangle = \langle \delta_0, f\,\varphi\rangle = f(0)\,\varphi(0) $$

• First interpretation of your problem. If $x$ is a constant (i.e. $f$ is the constant function $f(y) = \frac{1}{x}$) then for any continuous function $\varphi$ $$ \langle \tfrac{1}{x}\,\delta_0, \varphi\rangle = \tfrac{1}{x}\langle \delta_0, \varphi\rangle = \frac{\varphi(0)}{x} $$ which converges to $0$ when $x\to \infty$, so $\tfrac{1}{x}\,\delta_0(y)\underset{x\to\infty}\to 0$ in the sense of weak convergence for measures (but this is quite a stupid problem, any distribution multiplied by $0$ gives $0$).

• Second interpretation of your problem. You are looking at the distribution $\frac{\delta_0(x)}{x}$. This is not a well defined distribution or measure, there is no meaning to this object, I would say it is $\pm\infty$ or undefined.

• Third interpretation of your problem. You actually do not care about what is happening when $x$ is close to $0$, so we can look at $\frac{\delta_0(x)}{1+x}$. This is a well defined distribution (since $\frac{1}{1+x}$ is continuous at $0$) and if one takes a test function $\varphi_x$ compactly supported in the ball of center $x$ and radius $1$ then $$ \langle\tfrac{\delta_0(y)}{1+y},\varphi_x(y)\rangle_y = \varphi_x(0) $$ which is $0$ as soon as $x>1$, and so, $$ \langle\tfrac{\delta_0(y)}{1+y},\varphi_x(y)\rangle_y \underset{x\to\infty}\to 0. $$ Remark however that $\tfrac{\delta_0(y)}{1+y} = \delta_0$ so it was not very useful to multiply by this function here. The above result just tells us that $\langle\delta_0,\varphi_x\rangle \underset{x\to\infty}\to 0$ but this is trivial since $\delta_0$ is supported in $0$ (i.e. $\delta_0$ "is $0$ in all points except $0$") ...

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EDIT (since you edited your post):

• $\delta^2$ has no meaning (or $+\infty$)!
• If you remove the square: $∫_{-x/2}^{x/2} \delta(\mathrm{d}t) = 1$, so $$ \lim_{x\to\infty} \frac{∫_{-x/2}^{x/2} \delta(\mathrm{d}t)}{x} = \lim_{x\to\infty} \frac{1}{x} = 0 $$