Answer
$$
\lim _{x \rightarrow \infty} e^{x-x^2}=0.
$$
Work Step by Step
Since $ x^2-x $ is always positive for all $ x>0$, then we have
$$
\lim _{x \rightarrow \infty} e^{x-x^2}=\lim _{x \rightarrow \infty} e^{-(x^2-x)}=0.
$$
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
Published by
W. H. Freeman
ISBN 10:
1464125260
ISBN 13:
978-1-46412-526-3
Chapter 7 - Exponential Functions - 7.1 Derivative of f(x)=bx and the Number e - Exercises - Page 327: 14
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