Calculus: Early Transcendentals 8th Edition

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

Chapter 2 - Review - Exercises: 12

Answer

$$\lim\limits_{x\to3}\frac{\sqrt{x+6}-x}{x^3-3x^2}=-\frac{5}{54}$$

Work Step by Step

$$A=\lim\limits_{x\to3}\frac{\sqrt{x+6}-x}{x^3-3x^2}$$ Multiply both numerator and denominator by $\sqrt{x+6}+x$, we have: - Numerator: $(\sqrt{x+6}-x)(\sqrt{x+6}+x)=(x+6)-x^2=-(x^2-x-6)=-(x-3)(x+2)$ - Denominator: $(x^3-3x^2)(\sqrt{x+6}+x)=x^2(x-3)(\sqrt{x+6}+x)$ Therefore, $$A=\lim\limits_{x\to3}\frac{-(x-3)(x+2)}{x^2(x-3)(\sqrt{x+6}+x)}$$$$A=-\lim\limits_{x\to3}\frac{x+2}{x^2(\sqrt{x+6}+x)}$$$$A=-\frac{3+2}{3^2\times(\sqrt{3+6}+3)}$$$$A=-\frac{5}{54}$$
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