Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 2 - Limits - Review Exercises: 14

Answer

$\dfrac {3}{10}$

Work Step by Step

$\lim _{x\rightarrow 3}\dfrac {\sqrt {3x+16}-5}{x-3}=\lim _{x\rightarrow 3}\dfrac {\left( \sqrt {3x+16}-5\right) \left( \sqrt {3x+16}+5\right) }{\left( x-3\right) \left( \sqrt {3x+16}+5\right) }=\lim _{x\rightarrow 3}\dfrac {\left( \sqrt {3x+16}\right) ^{2}-5^2}{\left( x-3\right) \left( \sqrt {x+16}+5\right) }=\lim _{x\rightarrow 3}\dfrac {3x+16-25}{\left( x-3\right) \left( \sqrt {3x+16}+5\right) }=\lim _{x\rightarrow 3}\dfrac {3\left( x-3\right) }{\left( x-3\right) \left( \sqrt {3x+16}+5\right) }=\lim _{x\rightarrow 3}\dfrac {3}{\sqrt {3x+16}+5}=\dfrac {3}{\sqrt {3\times 3+16}+5}=\dfrac {3}{5+5}=\dfrac {3}{10}$
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