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 69: 88

Answer

We can guess here that $$\lim_{x\to3}\frac{x^2-9}{\sqrt{x^2+7}-4}=8$$

Work Step by Step

$$\lim_{x\to3}\frac{x^2-9}{\sqrt{x^2+7}-4}$$ (a) The graph is shown below. (b) Looking at the graph, as $x\to3$, we see that $\frac{x^2-9}{\sqrt{x^2+7}-4}$ approaches as close as $8$. Therefore, we can guess here that $$\lim_{x\to3}\frac{x^2-9}{\sqrt{x^2+7}-4}=8$$
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