limit 0 times infinity, rewrite to find the limit

Question

I need some help with: $\lim_{x\to 0+} x^3\cdot e^{1/x}$. How to start?

I've tried substitution $(y=1/x)$ without any luck.

I would prefer not to use L'Hopitals rule and apologizes for a bad title line.

Try writing this as $\frac{e^{1/x}}{(1/x)^3}$ and then using L'Hopitals rule — Mufasa, Oct 09 '13 at 20:17
I forgott to write that I prefer not to use L'Hopitals rule. It should be possible without it. — iveqy, Oct 09 '13 at 20:18

score 2 · Accepted Answer · answered Oct 09 '13 at 20:20

You can carry on with your substitution. In the case $y = 1/x$, then as $x \to 0^+$, $y \to \infty$, and you want to look at the limit of $$ \lim_{y\to\infty} (1/y^3) e^y = \lim_{y\to\infty} \frac{e^y}{y^3}. $$

If you know, for instance, that the exponential grows faster than any polynomial, you can avoid L'Hopital's rule.

score 1 · Answer 2 · answered Oct 09 '13 at 20:21

Substitute $x=\frac1t$. Then this is the same as $$\lim_{t\to+\infty}\frac{e^t}{t^3}$$ Using the wellknown inequality $e^x\ge 1+x$, we have $e^t=(e^{t/4})^4\ge (1+\frac t4)^4$, so for $t>0$ $$ \frac{e^t}{t^3}\ge\frac{(1+\frac t4)^4}{t^3}>\frac1{256}t.$$

(In short: Exponential always wins over polynomial)

score 0 · Answer 3 · answered Oct 09 '13 at 20:18

0

Hint: $e^{1/x} > \dfrac{1}{4!x^4}$.

answered Oct 09 '13 at 20:18

njguliyev

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Sorry, I don't get it... – iveqy Oct 09 '13 at 20:20
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$e^t=1+t+\dfrac{t^2}{2!}+\dfrac{t^3}{3!}+\dfrac{t^4}{4!}+\ldots$ – njguliyev Oct 09 '13 at 20:22

limit 0 times infinity, rewrite to find the limit

3 Answers3