Graphically, the limit is 0, but algebraically, why isn't it 1?

Question

$$\lim_{x\to\infty}e^{x-x^2} = 0$$

I'm a bit lost on this one. Graphically, it tends to 0, hence the limit is 0. But, how could I prove it algebraically?

Answer 1

Intuitively, $e^{-x^2}$ goes to zero very fast. If you know $\lim_{x \to \infty}e^{-x}=0$ you can just say that $x-x^2 \lt -x$ for $x \gt 2$, so $e^{x-x^2} \lt e^{-x} \to 0$.

Answer 2

$e^{x-x^2}=e^{x(1-x)}$. This means that when $x$ goes to $\infty$, the exponent goes to $\infty\cdot(-\infty)=-\infty$, and $e^{-\infty}=0$.

Answer 3

$$\lim_{x\to\infty}e^{x-x^2} = \lim_ {x\to\infty} \frac {e^x}{e^{x^2}} = \lim_ {x\to\infty} \frac {1}{e^{x^2-x}} = 1/{\infty} = 0 $$