I need to evaluate the limit

$$ \lim_{\epsilon \rightarrow 0+} \frac{1}{\epsilon}P\left(\frac{u}{\epsilon}\right),$$ where $P(u)$ is some continuous probability density function. Notably, this function and all of its derivatives should vanish as its argument goes to plus or minus infinity.

For physical reasons I suspect the limit may end up involving $\frac{d}{du} \sigma(u)P(u), $ where $\sigma(u)$ is the sign (signum) function (a generalized function).

This limit has a curious feature. Since $P(u)$ and all of its derivatives vanish at infinity, one can apply L'Hôpital's repeatedly, and the indeterminant form never resolves:

$$ \frac{P(u/\epsilon)}{\epsilon} = -u \frac{P'(u/\epsilon)}{\epsilon^2} = (-u)^2 \frac{P''(u/\epsilon)}{2\epsilon^3}=\dots = (-u)^n \frac{P^{(n)}(u/\epsilon)}{n!\epsilon^{n+1}}$$

Is there some way to use this property to derive a signum function representation of the limit?

Some exploratory stuff: The function $P(u/\epsilon - s u/\epsilon)$ where $s$ is some parameter near $0$ has Taylor expansion $$ P\left(\frac{u}{\epsilon}-s\frac{u}{\epsilon}\right) = \sum_{n=0}^\infty \frac{s^n}{n!}\Big(-\frac{u}{\epsilon}\Big)^nP^{(n)}\Big(\frac{u}{\epsilon}\Big), $$ which because of the above "curious feature" becomes $$ P\left(\frac{u}{\epsilon}-s\frac{u}{\epsilon}\right) = P(u/\epsilon)\sum_{n=0}^\infty s^n =\frac{P(u/\epsilon)}{1-s}. $$ This is odd, but I am not sure if it is useful.

  • $\begingroup$ If $x\mapsto 1/x$ then the problem is equivalent to find $\displaystyle \lim_{x\to +\infty}xf(ux)$ with $f$ a continuous pdf. $\endgroup$
    – A. P.
    Mar 11, 2022 at 0:39
  • 2
    $\begingroup$ The distributional limit exists and is the same for any $P$, since $$\lim_{\epsilon \to 0^+} \int_{\mathbb R} \frac 1 \epsilon P {\left( \frac u \epsilon \right)} \phi(u) \, du = \lim_{\epsilon \to 0^+} \int_{\mathbb R} P(u) \phi(\epsilon u) \, du = \int_{\mathbb R} \, \lim_{\epsilon \to 0^+} P(u) \phi(\epsilon u) \, du$$ as long as $\phi$ is continuous and bounded. $\endgroup$
    – Maxim
    Mar 11, 2022 at 14:36
  • $\begingroup$ Thanks for the suggestion @Maxim. Good to learn that it exists! I see that I should be investigating an integral over a test function to find a distributional limit. Is there a means to go further? What you wrote seems to suggest the distributional limit is $P(u)\phi(0 \pm)$, where $\pm = \text{sign}(u)$. Might this relate to $\frac{d}{du} \text{sign}(u) P(u)$? Can we place any assumptions on the test function? $\endgroup$ Mar 11, 2022 at 14:55

1 Answer 1


The limit may not exist. I am ignoring continuity in my example, but making it continuous is obvious For n > N. let $f(x)=\frac{1}{n^2}$ for $n^2-1\le x\le n^2$ and $=0$ otherwise. The limit as $x\to \infty$ does not exist.

  • $\begingroup$ That doesn't integrate to 1, did you make a typo in your answer? $\endgroup$
    – Clement C.
    Mar 11, 2022 at 6:50
  • $\begingroup$ @ClementC. I was not trying to gve an exact answer. To make it a probability density, each term has to be divided by the sum. The point of the example was to show that the limit may not exist at all - lower limit zero and upper limit > zero (exact value not important). $\endgroup$ Mar 12, 2022 at 4:20

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