Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 2: Limits and Continuity - Section 2.2 - Limit of a Function and Limit Laws - Exercises 2.2: 30

Answer

$-\dfrac{1}{2}$

Work Step by Step

We want to find $\lim\limits_{y \to 0}\dfrac{5y^3+8y^2}{3y^4-16y^2}$, but we can't use the quotient rule for limits because the limit of the denominator as $y$ approaches 0 is zero, and if we try to substitute $y=0$ directly, we get zero in both the numerator and the denominator. This means we have a common factor of $y$ in both the numerator and the denominator. (Actually, we have a common factor of $y^2$.) Note that $\dfrac{5y^3+8y^2}{3y^4-16y^2}=\dfrac{y^2(5y+8)}{y^2(3y^2-16)}=\dfrac{5y+8}{3y^2-16}$ for all $y \neq 0.$ Thus $\lim\limits_{y \to 0}\dfrac{5y^3+8y^2}{3y^4-16y^2}=\lim\limits_{y \to 0}\dfrac{5y+8}{3y^2-16}=\dfrac{5(0)+8}{3(0)^2-16}=\dfrac{8}{-16}=-\dfrac{1}{2}.$
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