Question about limit with exponents I tried to solve:
$$\lim\limits_{x \to \infty} e^{x-x^2}$$
but I can't get it done. I've tried to use Hopital's rule after rewriting it to:
$$\lim\limits_{x \to \infty} \frac{e^x}{e^{x^2}}$$
but this does not lead to a solition.
Anyone with a good hint?
 A: Note a polynomial is equivalent at infinity to its highest degree term, i.e. here: $\;x-x^2\sim_\infty -x^2$, hence $\lim_{x\to\infty}(x-x^2)=\lim_{x\to\infty}(-x^2)=-\infty$. By continuity $\;\mathrm e^{x-x^2}$ tends to $0$.
A: You can mind in this way: since
$$x^2 > x$$
as $x\to \infty$, then $x - x^2 = x(1-x) \approx x\cdot(-x) = -x^2$ hence
$$e^{x -x^2} \approx e^{-x^2} = 0$$
A: If $x > 2$ then $-x = x - 2x > x - x^2$ so $e^{x - x^2} < e^{-x}$ for $x > 2$. and $e^{x-x^2} > 0$ so $0 < \lim_{x\rightarrow \infty}e^{x - x^2}<  \lim_{x\rightarrow \infty}e^{-x}= 0$.  so $\lim_{x\rightarrow \infty}e^{x - x^2} = 0$.
... or ....
$\frac {de^x}{dx} = e^x$,  $\frac {de^{x^2}}{dx} = 2xe^{x^2} > 2x e^{x}$ for $x > 1$
So by l'hopitals rule $\lim \frac {e^x}{e^{x^2}} = \lim \frac {e^x}{2xe^2} \le \lim \frac 1{2x} = 0$.  So $e^{x - x^2} > 0$, $\lim e^{x- x^2} \ge 0$ so $\lim e^{x-x^2} = 0$.  
.... or ....
For $\epsilon > 0$ and $x > 1 + \sqrt{\ln 1/{\epsilon}}$ then $|e^{x - x^2} - 0|= 1/e^{x^2 - x} < 1/e^{(x - 1)^2} < 1/e^{\sqrt{\ln 1/\epsilon}^2}= \epsilon$.
....
or 
as $e^x$ is continuous $\lim e^{x- x^2} = e^{\lim x - x^2 = \lim -y} = 0$.
As long as you realize $x-x^2 \rightarrow -\infty$ and $\lim_{y\rightarrow -\infty} e^y = 0$ you have many options to prove it.
A: 
One way forward uses only the elementary inequality
$$\bbox[5px,border:2px solid #C0A000]{e^x\ge 1+x} \tag 1$$
along with the squeeze theorem.

Proceeding, we write
$$e^{x-x^2}=e^{-(x-1/2)^2}e^{1/4}=\frac{e^{1/4}}{e^{(x-1/2)^2}}$$
Then, applying $(1)$ reveals
$$0\le \frac{e^{1/4}}{e^{(x-1/2)^2}}\le \frac{e^{1/4}}{1+(x-1/2)^2}$$
whereupon application of the squeeze theorem yields the coveted limit
$$\lim_{x\to \infty}e^{x-x^2}=0$$
