Showing that an exponential function decays faster than a polynomial.

Question

Let $\delta>0$ be a small constant. I want to show that the function $$ f(x) = e^{-x^{-\delta}} $$

Is asymptotically bounded above by $x^k$ for any $k\in \mathbb{N}$. In other words: $$ \limsup_{x\rightarrow0}\frac{e^{-x^{-\delta}}}{x^k} = C $$

for some constant $C$.

Is there any technique besides L'hospital's that applies here?

L'hospital's rule seems to not help, since the chain rule for the top term gives us a $x^{-\delta-1}$ which ends up adding a $\delta$ to the exponent of the bottom term when all is said and done.

I have tried to bound $f$ above by some more reasonable function, but I can't seem to find one.

I also tried exploiting the Maclaurin series of $f$, but that again fails because the derivatives of $f$ are not defined at zero.

I am wondering if anyone here knows of a standard result or something that shows this is the case. Thank you

Answer 1

The expression can be rewritten as $\frac{1}{D}$ where $D=x^ke^{(\frac{1}{x})^\delta}$. Expand the exponential as a power series and get $D=\sum_{n=0}^\infty \frac{1}{n!}(\frac{1}{x})^{n\delta-k}$ The sum consists of positive terms, and for all $n\gt \frac{k}{\delta}$, the terms $\to \infty$ as $x\to 0$, so $\frac{1}{D}\to 0$. Therefore no positive $C$.

Answer 2

Claim: $\lim sup_{x \to 0+} \frac {e^{-x^{-\delta}}} {x^{k}}=0$. Let $\epsilon >0$ and $n >\frac k {\delta}$. Then $x^{k-n\delta} > \frac {n!} {\epsilon}$ for all $x$ sufficiently large. This gives $\frac {(x^{-\delta})^{n}} {n!} >\frac 1 {\epsilon} \frac 1 {x^{k}}$. Hence $e^{x^{-\delta}} >\frac 1 {\epsilon} \frac 1 {x^{k}}$. This is same as $e^{-x^{-\delta}}<\epsilon x^{k}$. This completes the proof.