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.