$\lim_{x \rightarrow 0^+} e^{-1/x}/x^3$

Question

Sorry for asking a possibly silly question. I was evaluating a limit, and in the end, it boiled down to showing that $\lim_{x \rightarrow 0^+} e^{-1/x}/x^3=0$.

All WolframAlpha says is

Hint: as $x\rightarrow0$, $e^{-1/x}/x^3$ approaches $0$,

which makes me worry that I'm either missing something simple, or that the limit is harder to prove than it seems.

Using l'Hôpital's rule increases the degree of the denominator, which doesn't seem very useful. I also found it difficult to group terms for an epsilon-delta proof. I additionally thought about squeezing it between $0$ and $x$, but I could not find a way to show that $$e^{-1/x}<x^4,$$ so that route also felt like a dead end.

How can this limit be found or proven?

Answer 1

Note that:$$\lim_{x\to0^+}\frac{e^{-1/x}}{x^3}=\lim_{x\to+\infty}\frac{e^{-x}}{\left(\frac1x\right)^3}=\lim_{x\to+\infty}\frac{x^3}{e^x}.$$Can you take it from here?

Answer 2

By Taylor formula, we know that $e^{u}>u^{4}$ for large $u>0$, then for $x>0$ small, $1/x>0$ is large, then $e^{-1/x}<x^{4}$, so $e^{-1/x}/x^{3}<x$ now use Squeeze Theorem to conclude that $\lim_{x\rightarrow 0^{+}}e^{-1/x}/x^{3}=0$.

Answer 3

Let $x\gt 0:$

$e^{1/x} =$

$1+ 1/x+(1/2!)(1/x)^2 +(1/3!)(1/x)^3 + (1/4!)(1/x)^4...\gt (1/4!)(1/x)^4$

implies:

$\dfrac{e^{-1/x}}{x^3} = \dfrac{1}{(e^{1/x})x^3} \lt$

$\dfrac {1}{(1/4!)(1/x)} =(4!)x$

Limit $x \rightarrow 0^+$ is?