Answer
$$\lim\limits_{x\to\infty}e^{x-x^2}=0$$
Work Step by Step
$$A=\lim\limits_{x\to\infty}e^{x-x^2}=\lim\limits_{x\to\infty}e^{x(1-x)}$$
As $x\to\infty$, $(1-x)$ approaches $-\infty$. That means $x(1-x)$ approaches $-\infty$.
Therefore, $$A=e^{-\infty}=\frac{1}{e^{\infty}}=0$$
