## Calculus (3rd Edition)

$$\lim _{x \rightarrow \infty} e^{x-x^2}=0.$$
Since $x^2-x$ is always positive for all $x>0$, then we have $$\lim _{x \rightarrow \infty} e^{x-x^2}=\lim _{x \rightarrow \infty} e^{-(x^2-x)}=0.$$