Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 2 - Limits - 2.5 Evaluating Limits Algebraically - Preliminary Questions - Page 72: 1


$$ \frac{x^{2}-1}{\sqrt{x+3}-2} $$

Work Step by Step

A rational function $f$ is indeterminate at $x=1$ if the formula for $f(1)$ yields un undefined expression of the type $\dfrac{\infty}{\infty}$ or $\dfrac{0}{0}$. We check $f(1)$ for each of the given functions. $1)$ $f(x)=\dfrac{x^2+1}{x-1}$ $f(1)=\dfrac{1^2+1}{1-1}=\dfrac{2}{0}$ (not indeterminate) $2)$ $f(x)=\dfrac{x^2-1}{x+2}$ $f(1)=\dfrac{1^2-1}{1+2}=\dfrac{0}{3}$ (not indeterminate) $3)$ $f(x)=\dfrac{x^2-1}{\sqrt{x+3}-2}$ $f(1)=\dfrac{1^2-1}{\sqrt{1+3}-2}=\dfrac{0}{0}$ (indeterminate) $4)$ $f(x)=\dfrac{x^2+1}{\sqrt{x+3}-2}$ $f(1)=\dfrac{1^2+1}{\sqrt{1+3}-2}=\dfrac{2}{0}$ (not indeterminate) So the function is $\dfrac{x^2-1}{\sqrt{x+3}-2}$ is the indeterminate one.
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