## Calculus 8th Edition

Published by Cengage

# Chapter 1 - Functions and Limits - 1.6 Calculating Limits Using the Limit Laws - 1.6 Exercises - Page 70: 7

390

#### Work Step by Step

$\displaystyle \lim_{x\rightarrow 8}(1+\sqrt[3]{x})(2-6x^{2}+x^{3})$= ...Law 4 (product ), $=\displaystyle \lim_{x\rightarrow 8}(1+\sqrt[3]{x})\cdot\lim_{x\rightarrow 8}(2-6x^{2}+x^{3})$ = ...Laws 1 (sum), 2 (difference), and 3 (constant multiple) $=(\displaystyle \lim_{x\rightarrow 8}1+\lim_{x\rightarrow 8}\sqrt[3]{x})\cdot(\lim_{x\rightarrow 8}2-6\lim_{x\rightarrow 8}x^{2}+\lim_{x\rightarrow 8}x^{3})$ = ...Laws 7:($\displaystyle \lim_{x\rightarrow a}c=c$), 10: ( $\displaystyle \lim_{x\rightarrow \mathit{0}}\sqrt[n]{x}=\sqrt[n]{a}$), and 9: $( \displaystyle \lim_{x\rightarrow a}x^{n}=a^{n})$, $=(1+\sqrt[3]{8})\cdot(2-6\cdot 8^{2}+8^{3})$ $=(3)(130)$ $=390$

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