Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am trying to solve a Limit using L'Hôpital's rule with $e^x$

So my question is how to find $$\lim_{x\rightarrow \infty} x^3 e^{-x^2}$$

I know to get upto this part here, but I'm lost after that

$$\lim_{x\rightarrow \infty} \frac{x^3}{e^{x^2}}$$

share|cite|improve this question
How does the 3x^2 become 3x then 3 on the numerator? – soniccool Nov 23 '12 at 22:59
Oh its a chain rulee – soniccool Nov 23 '12 at 23:00

\begin{align} \lim_{x \to \infty} \dfrac{x^3}{e^{x^2}}& = \lim_{x \to \infty} \dfrac{3x^2}{2xe^{x^2}} & \text{By L'Hôpital's rule.}\\ & = \lim_{x \to \infty} \dfrac{3x}{2e^{x^2}} & \text{Cancel of the $x$ in numerator and denominator.}\\ & = \lim_{x \to \infty} \dfrac{3}{4xe^{x^2}} & \text{By L'Hôpital's rule.}\\ & = 0 & \text{Since $x \to \infty$ and $\exp(x^2) \to \infty$ as $x \to \infty$.} \end{align}


Here is another way out. We have that $e^{x^2} = 1 + x^2 + \dfrac{x^4}{2!} + \mathcal{O}(x^6) \geq \dfrac{x^4}2$. Hence, we have that $$0 \leq \dfrac{x^3}{e^{x^2}} \leq \dfrac{x^3}{x^4/2} = \dfrac2x$$ Hence, $$0 \leq \lim_{x \to \infty} \dfrac{x^3}{e^{x^2}} \leq \lim_{x \to \infty} \dfrac2x = 0$$

share|cite|improve this answer

So now you take the derivative of the top and the bottom and get $$ \lim_{x\to \infty} \frac{3x^2}{2xe^{x^2}} = \lim_{x\to \infty} \frac{3x}{2e^{x^2}} $$ Taking derivatives again, you get $$ \lim_{x\to \infty}\frac{3}{4xe^{x^2}}. $$ I hope that you can find the limit from here.

share|cite|improve this answer

$$\begin{align} \lim_{x\rightarrow \infty} \dfrac{x^3}{e^{x^2}} &=\lim_{x\rightarrow \infty} \dfrac{3x^2}{e^{x^2}2x}\\ &=\lim_{x\rightarrow \infty} \dfrac{3x}{e^{x^2}2}\\ &=\lim_{x\rightarrow \infty} \dfrac{3}{e^{x^2}.2.2x}\\ &=0 \end{align}$$

share|cite|improve this answer
I edited your answer using the align environment. Hope it is ok... – Thomas Nov 23 '12 at 23:00
@Thomas, of course, I didn't use it since I was trying to be first :|. – Inquest Nov 24 '12 at 15:33

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.