Calculus: Early Transcendentals 8th Edition

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

Chapter 2 - Section 2.5 - Continuity - 2.5 Exercises - Page 124: 14


Prove that $\lim\limits_{x\to2}f(x)=f(2)$

Work Step by Step

*NOTES TO REMEMBER: $f(x)$ is continuous at $a$ if and only if $$\lim\limits_{x\to a}f(x)=f(a)$$ We consider $\lim\limits_{x\to2}f(x)$ $=\lim\limits_{x\to2}(3x^4-5x+\sqrt[3]{x^2+4})$ $=3\lim\limits_{x\to2}x^4-5\lim\limits_{x\to2}x+\lim\limits_{x\to2}\sqrt[3]{x^2+4}$ $=3\lim\limits_{x\to2}x^4-5\lim\limits_{x\to2}x+\sqrt[3]{\lim\limits_{x\to2}(x^2+4)}$ $=3\lim\limits_{x\to2}x^4-5\lim\limits_{x\to2}x+\sqrt[3]{\lim\limits_{x\to2}x^2+\lim\limits_{x\to2}4}$ $=3\times2^4-5\times2+\sqrt[3]{2^2+4}$ $=f(2)$ Therefore, $f(x)$ is continuous at $2$.
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