Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 2: Limits and Continuity - Section 2.2 - Limit of a Function and Limit Laws - Exercises 2.2 - Page 57: 63



Work Step by Step

Here, we have $\lim\limits_{x\to0}\sqrt{5-2x^2}=\sqrt{5-2\times0^2}=\sqrt{5-0}=\sqrt5$ and $\lim\limits_{x\to0}\sqrt{5-x^2}=\sqrt{5-0^2}=\sqrt5$ Then, $\lim\limits_{x\to0}\sqrt{5-2x^2}=\lim\limits_{x\to0}\sqrt{5-x^2}=\sqrt5$ As we can see that $\sqrt{5-2x^2}\le f(x)\le \sqrt{5-x^2}$ for $-1\le x\le 1$ Thus, as per Sandwich Theorem, we can see that $\lim\limits_{x\to0}f(x)=\sqrt5$
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