Solving $\lim_{x\to-\infty}x^2\cdot e^x$ with L'Hopital 
Use the L'Hopital rule to solve:
$$\lim_{x\to-\infty}x^2\cdot e^x$$

I need a quotient of infinities or of zeroes. One way could be this:
$$\frac{e^x}{x^{-2}} = \frac{0}{0}$$
So we apply the L'Hopital rule:
$$\frac{e^x}{-2x^{-3}}$$
Oh, this is not going to work. We'll be applying L'Hopital countless times without luck.
Perhaps this arrangement will do:
$$\frac{x^{2}}{e^{-x}} = \frac{\infty}{0}$$
Nope.
Maybe I could use one of the properties of natural logarithms. Since $e^x = \ln(x)$ I could have
$$\frac{x^2}{\ln(-x)} = \frac{\infty}{\infty}$$
Great! Now we can apply the L'Hopital rule:
$$\frac{2x}{\frac{-1}{\ln(-x)}} = \frac{-\infty}{0}$$
Dammit. Well, maybe we can re-arrange this:
$$\frac{2x}{\frac{-1}{\ln(-x)}} = 2x\cdot -\ln(-x)$$
Then,
$$2x\cdot -\ln(-x) = -\infty \cdot -\infty = \infty$$
Apparently this is wrong. The answer should be $0$.
What was my mistake?
 A: The nope should be a yeah! 
$$
\lim_{x\to-\infty}e^{-x} = \lim_{x\to \infty}e^{x} = \infty
$$
A: $$\lim_{x\to-\infty}x^2\cdot e^x=\lim_{x\to\infty}\frac{x^2}{ e^x}$$
A: $$\lim_{x\to -\infty}e^xx^2=$$
$$\lim_{x\to -\infty}\frac{x^2}{e^{-x}}=$$
$$\lim_{x\to -\infty}\frac{\frac{\text{d}}{\text{d}x}x^2}{\frac{\text{d}}{\text{d}x}e^{-x}}=$$
$$\lim_{x\to -\infty}\frac{2x}{-e^{-x}}=$$
$$\lim_{x\to -\infty}-2e^xx=$$
$$-2\left(\lim_{x\to -\infty}e^xx\right)=$$
$$-2\left(\lim_{x\to -\infty}\frac{x}{e^{-x}}\right)=$$
$$-2\left(\lim_{x\to -\infty}\frac{\frac{\text{d}}{\text{d}x}x}{\frac{\text{d}}{\text{d}x}e^{-x}}\right)=$$
$$-2\left(\lim_{x\to -\infty}\frac{1}{-e^{-x}}\right)=$$
$$-2\left(\lim_{x\to -\infty}-e^x\right)=$$
$$2\left(\lim_{x\to -\infty}e^x\right)=$$
$$2\left(\exp\left(\lim_{x\to -\infty}x\right)\right)=0$$
A: $ln(e^x) = x$, not $ln(x) =e^x$
And indeed, the limit is $0$. 
You have $x^2 e^x = \frac{x^2}{e^{-x}}$,  and, by applying l'Hôpital's rule twice, you get :
$\lim_{x\rightarrow-\infty} x^2 e^x = \lim_{x\rightarrow-\infty} \frac{2x} {-e^{-x}} = \lim_{x\rightarrow-\infty} \frac{2} {e^{-x}} = \lim_{x\rightarrow-\infty} 2 e^x = 0 $
