Limit using L'Hôpital's rule

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

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

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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}

EDIT

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, $$0 \leq \lim_{x \to \infty} \dfrac{x^3}{e^{x^2}} \leq \lim_{x \to \infty} \dfrac2x = 0$$

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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.

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\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}

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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