Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 11 - Section 11.2 - Finding Limits Using Properties of Limits - Exercise Set - Page 1153: 41


$ \dfrac{1}{3}$

Work Step by Step

Recall that if $ f $ is a polynomial function, then we have $\lim_\limits{x\to a}f(x)=f(a)$. In order to to find the limit, we will plug $ a $ into the function and then simplify. $\lim_\limits{x\to 2} \dfrac{\dfrac{x^2-4}{x^3-8}-\dfrac{1}{4}}{x}=\lim_\limits{x\to 2} \dfrac{(x-2)(x+2)}{(x-2)(x^2+2x+4)}$ or, $= \lim_\limits{x\to 2} \dfrac{x+2}{x^2+2x+4}$ or, $=\dfrac{2+2}{(2)^2+2(2)+4}$ or, $= \dfrac{1}{3}$
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