Let function $f$ that depends on parameter $h>0$ with $$f(x;h)=x e^{-\frac{x}{h}}, x>0.$$
I want to show the following:
For fixed $x>0$ and for each $n \in \mathbb{N}$ we have $f(x;h)<<h^n,h \to 0^{+}$ and
$\max_{x >0} f(x;h)=O(h), h \to 0^{+}$.
I have thought the following:
$$\lim_{h \to 0^{+}} \frac{f(x,h)}{h^n}=\lim_{h \to 0^{+}} \frac{x e^{-\frac{x}{h}}}{h^n}\overset{DLH}{=}\frac{x^2}{n} \lim_{h \to 0^+}\frac{e^{-\frac{x}{h}}}{h^{n+1}}= \dots=\frac{x^3}{n+1} \lim_{h \to 0^+}\frac{e^{-\frac{x}{h}}}{h^{n+2}}$$
Is this right so far? But can this fornula help? The limit will always have this form so I think that we cannot evaluate it like that.
Do we show this in an other way?