Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 6 - Inverse Functions - 6.8 Indeterminate Forms and I'Hospital's Rule - 6.8 Exercises - Page 500: 12



Work Step by Step

Given: $\lim\limits_{x\to1/2}\dfrac{6x^2+5x-4}{4x^2+16x-9}$ This shows an indeterminate form of type $\dfrac{0}{0}$, and so we will apply L'Hospital Rule. $\lim\limits_{x\to1/2}\dfrac{\dfrac{d}{dx}(6x^2+5x-4)}{\dfrac{d}{dx}(4x^2+16x-9)}=\lim\limits_{x\to1/2}\dfrac{12x+5}{8x+16}\\=\dfrac{12\times\dfrac{1}{2}+5}{8\times\dfrac{1}{2}+16}\\=\dfrac{6+5}{4+16}=\dfrac{11}{20}$
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