## Calculus 8th Edition

Published by Cengage

# Chapter 1 - Functions and Limits - 1.6 Calculating Limits Using the Limit Laws - 1.6 Exercises - Page 70: 8

#### Answer

$\displaystyle \frac{4}{49}$

#### Work Step by Step

$\displaystyle \lim_{t\rightarrow 2}(\frac{t^{2}-2}{t^{3}-3t+5})^{2}$= ... Law 6, $\displaystyle \lim_{x\rightarrow a}[f(x)]^{n}=[\lim_{x\rightarrow a}f(x)]^{n}$ $=(\displaystyle \lim_{t\rightarrow 2}\frac{t^{2}-2}{t^{3}-3t+5})^{2}$ ... Law 5, (quotient) $=(\displaystyle \frac{\lim_{t\rightarrow 2}(t^{2}-2)}{\lim_{t\rightarrow 2}(t^{3}-3t+5)})^{2}$ ...Laws 1 (sum), 2 (difference), and 3 (constant multiple) $=(\displaystyle \frac{\lim_{t\rightarrow 2}t^{2}-\lim_{t\rightarrow 2}2}{\lim_{t\rightarrow 2}t^{3}-3\lim_{t\rightarrow 2}t+\lim_{t\rightarrow 2}5})^{2}$ ... Laws 9: $( \displaystyle \lim_{x\rightarrow a}x^{n}=a^{n})$,7:($\displaystyle \lim_{x\rightarrow a}c=c$), and 8: ( $\displaystyle \lim_{x\rightarrow a}x=a$) $=(\displaystyle \frac{4-2}{8-3(2)+5})^{2}$ $=(\displaystyle \frac{2}{7})^{2}$ $=\displaystyle \frac{4}{49}$

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