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: 48

Answer

$10$

Work Step by Step

$\lim _{t\rightarrow 3}\left( \left( 4t-\dfrac {2}{t-3}\right) \left( 6+t-t^{2}\right) \right) =\dfrac {4t\left( t-3\right) -2}{\left( t-3\right) }\times \left( 9-3+t-t^{2}\right) =\dfrac {4t\left( t-3\right) -2}{t-3}\left( 3^{2}-t^{2}-\left( 3-t\right) \right) =\dfrac {4t\left( t-3\right) -2}{t-3}\times (\left( 3-t\right) \left( 3+t\right) -\left( 3-t\right) ) =\dfrac {4t\left( t-3\right) -2}{\left( t-3\right) }\left( \left( 3-t\right) \times \left( 3+t-1\right) \right) =\dfrac {4t\left( t-3\right) -2}{t-3}\times \left( 3-t\right) \left( 2+t\right) =\dfrac {4t\left( t-3\right) -2}{t-3}\times \left(-( t-3\right) \left( 2+t\right) ) =-\left( 4t\left( t-3\right) -2\right) \left( 2+t\right) =-\left( 4\times 3\left( 3-3\right) -2\right) \times \left( 2+3\right) =2\times 5=10 $
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