Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 4 - Section 4.4 - Indeterminate Forms and l''Hospital''s Rule - 4.4 Exercises - Page 316: 23

Answer

$ - \frac{1}{3}$

Work Step by Step

$$\eqalign{ & \mathop {\lim }\limits_{x \to 3} \frac{{\ln \left( {x/3} \right)}}{{3 - x}} \cr & {\text{Use the quotient property of limits}} \cr & = \frac{{\mathop {\lim }\limits_{x \to 3} \left[ {\ln \left( {x/3} \right)} \right]}}{{\mathop {\lim }\limits_{x \to 3} \left( {3 - x} \right)}} \cr & {\text{Evaluate the limit when }}x \to 3 \cr & = \frac{{\ln \left( {3/3} \right)}}{{3 - 3}} = \frac{0}{0} \cr & {\text{The limit is an indeterminate form type }}\frac{0}{0},{\text{ so we can apply}} \cr & {\text{the L'Hospital's Rule}}{\text{.}} \cr & \mathop {\lim }\limits_{x \to 3} \frac{{\ln \left( {x/3} \right)}}{{3 - x}} = \mathop {\lim }\limits_{x \to 3} \frac{{\frac{d}{{dx}}\left[ {\ln \left( {x/3} \right)} \right]}}{{\frac{d}{{dx}}\left[ {3 - x} \right]}} \cr & = \mathop {\lim }\limits_{x \to 3} \frac{{\frac{{1/3}}{{x/3}}}}{{ - 1}} \cr & = - \mathop {\lim }\limits_{x \to 3} \frac{1}{x} \cr & {\text{Evaluate the limit when }}x \to 3 \cr & = - \frac{1}{3} \cr & {\text{Therefore}}{\text{,}} \cr & \mathop {\lim }\limits_{x \to 3} \frac{{\ln \left( {x/3} \right)}}{{3 - x}} = - \frac{1}{3} \cr} $$
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