$\lim_{x \to 0} \frac{\exp[-x^{-\gamma}]}{x^d}=0$

Question

Trying to show for $\gamma \in (0,1]$ and some natural number $d$ that $$\lim_{x \to 0} \frac{\exp[-x^{-\gamma}]}{x^d}=0$$.

I've tried setting $t=x^{-\gamma}$ and you get for the right side limit

$$\lim_{t \to \infty} e^{-t}t(1/t)^{\frac{\gamma-d}{\gamma}} = \lim_{t \to \infty} e^{-t}t^{1-\frac{\gamma-d}{\gamma}} = \lim_{t \to \infty}$$

To show $t^{d/\gamma} / e^t \to 0$, recall that exponential growth is faster than polynomial growth, e.g. see here. — angryavian, Apr 05 '24 at 00:19
@angryavian perfect! does this also work for the left side limit? — mather, Apr 05 '24 at 19:11

score 1 · Answer 1 · answered Apr 05 '24 at 19:26

As mentioned in the comments, after your substitution, you get $$\text{your limit}=\lim_{t\to \infty} \frac{t^{\frac{d}{\gamma}}}{e^t}= 0$$ by a well-known result (see the link in the comments for a good reference).

You seem to be confused with right/left limits, but when $x \to 0$, $t \to \infty$, so this is a straightforward consequence of change of variable theorem for limits and you don't need to split the limits into left and right hand side limits.

$\lim_{x \to 0} \frac{\exp[-x^{-\gamma}]}{x^d}=0$

1 Answers1