Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 4 - Applications of the Derivative - 4.7 L'Hopital's Rule - 4.7 Exercises: 90

Answer

The solution is $$\lim_{x\to\infty}(x^2e^{1/x}-x^2-x)=\frac{1}{2}.$$

Work Step by Step

To solve this limit we will introduce a substitution $t=\frac{1}{x}$. Then also $x=\frac{1}{t}$ and when $x\to\infty$ then $t\to0^+$. Also we will use L'Hopital's rule. LR will stand for Apply L'Hopital's rule: $$\lim_{x\to\infty}(x^2e^{1/x}-x^2-x)=\lim_{t\to0^+}\left(\left(\frac{1}{t}\right)^2e^t-\left(\frac{1}{t}\right)^2-\frac{1}{t}\right)=\lim_{t\to0^+}\frac{e^t-1-t}{t^2}=\left[\frac{e^{0^+}-1-0^+}{(0^+)^2}\right]=\left[\frac{1-1-0}{0}\right]=\left[\frac{0}{0}\right][\text{LR}]=\lim_{t\to0^+}\frac{(e^t-1-t)'}{(t^2)'}=\lim_{t\to0^+}\frac{e^t-1}{2t}=\frac{1}{2}\lim_{t\to0^+}\frac{e^t-1}{t}=\frac{1}{2}\cdot1=\frac{1}{2},$$ where we used that $\lim_{t\to0^+}\frac{e^t-1}{t}=1$ which is well known and ready-to-use result.
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