University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 2 - Section 2.2 - Limit of a Function and Limit Laws - Exercises - Page 68: 67

Answer

$$\lim_{x\to-3}f(x)=-6$$
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Work Step by Step

$$f(x)=\frac{x^2-9}{x+3}$$ (a) The table is shown below. As we can see from 2 tables, as $x$ approaches $-3$, the value of $f(x)$ will get closer and closer to $-6$. So a logical estimate would be $\lim_{x\to-3}f(x)=-6$. (b) The graph is shown below. Again, looking at the graph, the closer $x$ approaches $-3$, the closer $y$ approaches $-6$. So we ca estimate $\lim_{x\to-3}f(x)=-6$. (c) $$\lim_{x\to-3}f(x)=\lim_{x\to-3}\frac{x^2-9}{x+3}=\lim_{x\to-3}\frac{(x-3)(x+3)}{x+3}$$ $$=\lim_{x\to-3}x-3=-3-3=-6$$ Therefore, $$\lim_{x\to-3}f(x)=-6$$
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