Calculus: Early Transcendentals (2nd Edition)

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

Chapter 2 - Limits - 2.3 Techniques for Computing Limits - 2.3 Exercises: 62

Answer

$-\dfrac {8}{225} $

Work Step by Step

$\lim _{x\rightarrow 3}\dfrac {\dfrac {1}{x^{2}+2x}-\dfrac {1}{15}}{x-3}=\dfrac {15-x^{2}-2x}{15\left( x^{2}+2x\right) \times \left( x-3\right) }=\dfrac {9+6-x^{2}-2x}{15\left( x^{2}+2x\right) \left( x-3\right) }=\dfrac {3^{2}+6-x^{2}-2x}{15\left( x^{2}+2x\right) \left( x-3\right) }=\dfrac {\left( 3-x\right) \left( 3+x\right) +\left( 6-2x\right) }{15\left( x^{2}+2x\right) \left( x-3\right) } =\dfrac {\left( 3-x\right) \left( 3+x\right) +2\left( 3-x\right) }{15\left( x-3\right) \left( x^{2}+2x\right) }=\dfrac {\left( 3-x\right) \left( 3+x+2\right) }{15\left( x-3\right) \left( x^{2}+2x\right) }=\dfrac {-\left( 5+x\right) }{15\left( x^{2}+2x\right) }=\dfrac {-\left( 5+3\right) }{15\left( 3^{2}+6\right) }=\dfrac {-8}{225} $
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.