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: 10



Work Step by Step

Consider: $\lim\limits_{t \to 1}f(t)=\lim\limits_{t \to 1}\dfrac{t^3-1}{4t^3-t-3}$ Need to check that the limit has an indeterminate form. Thus, $f(1)=\dfrac{1^3-1}{4(1)^3-1-3}=\dfrac{0}{0}$ 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: $\lim\limits_{t \to 1}\dfrac{3t^2}{12t^2-1}=\dfrac{3(1)^2}{12(1)^2-1}=\dfrac{3}{11}$
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