Finding asymptotes of $(2-x^2)/(e^x)$

I was trying to solve some exam question on calculus 1, and i found this "Sketch the graph of $(2-x^2)/(e^x)$"

I'm interested to find Horizontal Asymptotes of the graph.

1) when x approaches $+\infty$, $y = 0$.
2) when x approaches $-\infty$ i get the form $[{-\infty \over 0}]$ and i'm not sure what to do next, what does it mean $[{-\infty \over 0}]$ and what is the bast way to solve it.

Here is what i'm thinking, $-x^2$ will go infinity faster than $e^x$ so it look likes $(2-x^2)/(e^x)$ go to $-\infty$ as x approaches $-\infty$.

from that i'm kinda thinking $-\infty \over 0$=$-\infty$ or something like $(-1)\infty$* $1 \over 0$=$-\infty$

one more thing i will appreciate if someone give me a hint how to solve for slant asymptote too. i only knows how to do it with functions of Polynomial Quotient (long division)

The answers are correct, but keep in mind what "limit" means: you are approximating the behavior of you function for highly negative values, therefore as $x^2$ becomes bigger and bigger, $e^x$ gets very small positive values.
What do you get by dividing something big by something quite small? So you see that $$\frac{-\infty}{0^+}$$ is no indefinite form!
As for slant asymptote you have to check if $$\lim_{x\to \pm\infty}\frac{f(x)}{x}$$ exists and is finite. If so, you get your asymptote: $$y=mx+q;\ m= \lim_{x\to \pm\infty}\frac{f(x)}{x};\ q= \lim_{x\to \pm\infty} (f(x)-mx).$$
You are on the right track. Just remember that $e^x$ grows faster than $x^n$, however large $n$ is. This will give you the asymptotes of the curve.
As to your second question about slant asymptotes, there is a plethora of high school math sites where you can get help. I rather like $www.slideshare.net/nsimmons$