## Calculus: Early Transcendentals 8th Edition

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