Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 2 - Section 2.3 - Calculating Limits Using the Limit Laws - 2.3 Exercises - Page 103: 36

Answer

Squeeze Theorem $lim _{x\to 0}\left(\sqrt{x^3+x^2}\right)sin\frac{\pi }{x}=0$

Work Step by Step

Squeeze Theorem If $ h(x)≤f(x)≤g(x)$ If $\lim _{x\to a}h\left(x\right)=L$ and $\lim _{x\to a}g\left(x\right)=L$ then $\lim _{x\to a}f\left(x\right)=L$ We know $-1\le \frac{sin\pi }{x}\le 1$ Multiply by $\sqrt{x^3+x^2}$ $-\sqrt{x^3+x^2}\le \frac{\sqrt{x^3+x^2}sin\pi }{x}\le \sqrt{x^3+x^2}$ Since $\lim _{x\to 0}-\sqrt{x^3+x^2}=0$ $\lim _{x\to 0}\sqrt{x^3+x^2}=0$ Squeeze Theorem $lim _{x\to 0}\left(\sqrt{x^3+x^2}\right)sin\frac{\pi }{x}=0$ See graph.
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