Limits at infinity with exponentials I have never come across this type of problem so far and am lost on how to proceed. I was told never to plug in infinity because it can get theoretical. $\infty-\infty=0$? $e^0=0$?
Problem: $$\lim_{x\to \infty}e^{x-x^2}$$
 A: Note that $\lim_{x \to \infty} x - x^2 = -\infty$ so if we set $u = x - x^2$ we have
$$ \lim_{x \to \infty} e^{x - x^2} = \lim_{u \to -\infty} e^u = 0. $$
A: When evaluating a limit, you do not, strictly speaking, "plug in infinity" (although you'll be using your intuition to do so).  You are correct.  Consider the following:
$$lim_{x \rightarrow \infty}(x-x^2)$$
In this case, we see that the limit is $-\infty$.  Then, $e^{-\infty}$ = 0, as you specified.
A: ${lim_{x \rightarrow +\infty}x-x^2}=lim_{x \rightarrow +\infty}(-x^2)=-\infty$
$lim_{x \rightarrow +\infty}e^{x-x^2}=e^{-\infty}=0$
A: $ \lim_{x\to \infty}e^{x-x^2}=\lim_{x \to \infty}\frac{e^x}{e^{x^2}}=\lim_{x \to \infty}\frac{1}{e^{x^2-x}}=\lim_{x \to \infty}\frac{1}{e^{x(x-1)}}=\frac{1}{\infty}=0$
A: Using the inequality $e^x\ge 1+x$, which I showed in THIS ANSWER using only the limit definition of the exponential function and Bernoulli's Inequality, we can write
$$0\le e^{x-x^2}=\frac{1}{e^{x^2-x}}\le \frac{1}{1+x^2-x}$$
whereupon applying the squeeze theorem yields the coveted limit
$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to \infty}e^{x-x^2}=0}$$
