Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 4 - Section 4.4 - Indeterminate Forms and l''Hospital''s Rule - 4.4 Exercises: 8

Answer

$$\lim_{x\to3}\frac{x-3}{x^2-9}=\frac{1}{6}$$

Work Step by Step

$$A=\lim_{x\to3}\frac{x-3}{x^2-9}$$ In this exercise, the L'Hospital's Rule is actually not necessary. However, for the sake of everybody, I will carry out both methods, elementary one and L'Hospital's Rule one. 1) Method 1: Elementary method $$A=\lim_{x\to3}\frac{x-3}{x^2-9}$$ $$A=\lim_{x\to3}\frac{x-3}{(x-3)(x+3)}$$ $$A=\lim_{x\to3}\frac{1}{x+3}$$ $$A=\frac{1}{3+3}$$ $$A=\frac{1}{6}$$ 2) Method 2: L'Hospital's Rule Since $\lim_{x\to3}(x-3)=3-3=0$ and $\lim_{x\to3}(x^2-9)=3^2-9=0$, this limit is an indeterminate form of type $\frac{0}{0}$, and we can apply L'Hospital Rule $$A=\lim_{x\to3}\frac{x-3}{x^2-9}$$ $$A=\lim_{x\to3}\frac{\frac{d(x-3)}{dx}}{\frac{d(x^2-9)}{dx}}$$ $$A=\lim_{x\to3}\frac{1}{2x}$$ $$A=\frac{1}{2\times3}$$ $$A=\frac{1}{6}$$
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