Thomas' Calculus 13th Edition

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

Chapter 7: Transcendental Functions - Section 7.5 - Indeterminate Forms and L'Hopital's Rule - Exercises 7.5 - Page 409: 9

Answer

$-\dfrac{23}{7}$

Work Step by Step

Consider: $\lim\limits_{t \to -3}f(t)=\lim\limits_{t \to -3}\dfrac{t^3-4t +15}{t^2-t-12}$ Need to check that the limit has an indeterminate form. Thus, $f(-3)=\dfrac{(-3)^3-4(-3)+15}{(-3)^2-(-3)-12}=\dfrac{-27+12+15}{9+3-12}=\dfrac{0}{0}$ Now, apply L-Hospital's rule such as: $\lim\limits_{a \to b}f(x)=\lim\limits_{a \to b}\dfrac{g'(x)}{h'(x)}$ This implies: $\lim\limits_{t \to -3}\dfrac{3t^2-4}{2t-1}=\dfrac{3(-3)^2 -4}{2(-3)-1}=-\dfrac{23}{7}$
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