University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 2 - Section 2.2 - Limit of a Function and Limit Laws - Exercises - Page 67: 63



Work Step by Step

- Calculate $\lim_{x\to0}\sqrt{5-2x^2}$ and $\lim_{x\to0}\sqrt{5-x^2}$ $\lim_{x\to0}\sqrt{5-2x^2}=\sqrt{5-2\times0^2}=\sqrt{5-0}=\sqrt5$ $\lim_{x\to0}\sqrt{5-x^2}=\sqrt{5-0^2}=\sqrt5$ So, $\lim_{x\to0}\sqrt{5-2x^2}=\lim_{x\to0}\sqrt{5-x^2}=\sqrt5$ - Yet $\sqrt{5-2x^2}\le f(x)\le \sqrt{5-x^2}$ for $-1\le x\le 1$ Therefore, applying the Sandwich Theorem, we conclude that $$\lim_{x\to0}f(x)=\sqrt5$$
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