University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 2 - Practice Exercises - Page 111: 45



Work Step by Step

*Remember that $\lim_{x\to\pm\infty}\frac{a}{x^n}=a\lim_{x\to\pm\infty}\frac{1}{x^n}=a\times0=0$ $$A=\lim_{x\to-\infty}\frac{x^2-7x}{x+1}$$ Divide both numerator and denominator by the highest degree of $x$ in the denominator, which is $x$ in this case: $$A=\lim_{x\to-\infty}\frac{x-7}{1+\frac{1}{x}}$$ $$A=\frac{\lim_{x\to-\infty}(x-7)}{1+0}=\lim_{x\to-\infty}(x-7)$$ As $x\to-\infty$, $(x-7)$ approaches $-\infty$. Therefore, $$A=-\infty$$
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