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 - Page 56: 18

Answer

$\dfrac{1}{5}$

Work Step by Step

We know if $f=\dfrac{p}{q}$ is a rational function and $a$ is any real number, then $\lim\limits_{x \to a}f(x)=f(a)$ provided $q(a)$ is not zero. Hence since $\dfrac{y+2}{y^2+5y+6}$ is a rational function and $y^2+5y+6 \neq 0$ for $y=2$, we have $\lim\limits_{y \to 2}\dfrac{y+2}{y^2+5y+6}=\dfrac{2+2}{2^2+5(2)+6}=\dfrac{4}{4+10+6}=\dfrac{4}{20}=\dfrac{1}{5}.$
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