Answer
0
Work Step by Step
This limit is of the indeterminate form $\infty\cdot0$. Hence we use L'Hopital's: $\displaystyle\lim_{x\to\infty}\frac{x}{e^{x^2}}=\lim_{x\to\infty}\frac{1}{2xe^{x^2}}=\fbox{$0$}$.
Calculus 10th Edition
by Larson, Ron; Edwards, Bruce H.
Published by
Brooks Cole
ISBN 10:
1-28505-709-0
ISBN 13:
978-1-28505-709-5
Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - Review Exercises - Page 580: 76
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