Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 1 - Functions and Limits - 1.6 Calculating Limits Using the Limit Laws - 1.6 Exercises: 1

Answer

(a) -6 (b) -8 (c) 2 (d) -6 (e) does not exist because the denominator is 0 (f) 0

Work Step by Step

It is given that $\lim\limits_{x \to 2}$$f(x)=4$ $\lim\limits_{x \to 2}$$g(x)=-2$ $\lim\limits_{x \to 2}$$h(x)=0$ (a) $\lim\limits_{x \to 2}$ $[f(x) + 5g(x)]$ Apply the addition law and the constant law. $\lim\limits_{x \to 2}$$f(x) + 5$$\lim\limits_{x \to 2}$$g(x)$ Substitute. $(4) + 5(-2)$ Clean up and simplify $4-10$ $$-6$$ (b) $\lim\limits_{x \to 2}$$[g(x)]^3$ Apply the power law. $(\lim\limits_{x \to 2}g(x))^3$ Substitute. $(-2)^3$ $$-8$$ (c) $\lim\limits_{x \to 2}\sqrt {f(x)}$ Apply root law. $\sqrt {\lim\limits_{x \to 2}f(x)}$ Substitute. $\sqrt 4$ $$2$$ (d) $\lim\limits_{x \to 2}\frac{3f(x)}{g(x)}$ Apply the division law and the constant law. $\frac{3\lim\limits_{x \to 2}f(x)}{\lim\limits_{x \to 2}g(x)}$ Substitute. $\frac{3(4)}{(-2)}$ $$-6$$ (e) $\lim\limits_{x \to 2}\frac{g(x)}{h(x)}$ Apply the division law. $\frac{\lim\limits_{x \to 2}g(x)}{\lim\limits_{x \to 2}h(x)}$ Substitute. $\frac{-2}{0}$ The limit does not exist because the denominator is 0. (f)$\lim\limits_{x \to 2}\frac{g(x)h(x)}{f(x)}$ Apply the multiplication law and the division law. $\frac{\lim\limits_{x \to 2}g(x)*\lim\limits_{x \to 2}h(x)}{\lim\limits_{x \to 2}f(x)}$ Substitute. $\frac{(-2)(0)}{(4)}$ Simplify. $\frac{0}{4}$ $$0$$
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