## Calculus 8th Edition

Published by Cengage

# Chapter 1 - Functions and Limits - 1.6 Calculating Limits Using the Limit Laws - 1.6 Exercises - Page 70: 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$\$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.