Limit $\lim_{x\rightarrow \infty} x^3 e^{-x^2}$ 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
$$\lim_{x\rightarrow \infty} \frac{x^3}{e^{x^2}}$$
 A: \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, $$0 \leq \lim_{x \to \infty} \dfrac{x^3}{e^{x^2}} \leq \lim_{x \to \infty} \dfrac2x = 0$$
A: $$\begin{align}
\lim_{x\rightarrow \infty} \dfrac{x^3}{e^{x^2}}
&=\lim_{x\rightarrow \infty} \dfrac{3x^2}{e^{x^2}2x}\\
&=\lim_{x\rightarrow \infty} \dfrac{3x}{e^{x^2}2}\\
&=\lim_{x\rightarrow \infty} \dfrac{3}{e^{x^2}.2.2x}\\
&=0
\end{align}$$
A: 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.
