Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 2 - Section 2.6 - Limits at Infinity; Horizontal Asymptotes - 2.6 Exercises - Page 138: 43

Answer

(a) (i) $\lim\limits_{x \to 0^+}f(x) = 0$ (ii) $\lim\limits_{x \to 1^-}f(x) = -\infty$ (iii) $\lim\limits_{x \to 1^-}f(x) = \infty$ (b) Since the values seem to increase without bound, we could guess that: $\lim\limits_{x \to \infty}f(x) = \infty$ (c) We can see a sketch of the graph below.

Work Step by Step

(a) $f(x) = \frac{x}{ln~x}$ (i) As $~~x \to 0^+~~$, the value of $x$ is a very small positive number, while $ln~x$ approaches $-\infty$ $\lim\limits_{x \to 0^+}f(x) = 0$ (ii) As $~~x \to 1^-~~$, the value of $x$ is just less than 1, while $ln~x$ is a very small negative number. $\lim\limits_{x \to 1^-}f(x) = -\infty$ (iii) As $~~x \to 1^+~~$, the value of $x$ is a little more than 1, while $ln~x$ is a very small positive number. $\lim\limits_{x \to 1^-}f(x) = \infty$ (b) We can evaluate $f(x)$ for increasing values of $x$: $f(10) = \frac{10}{ln~10} = 4.34$ $f(100) = \frac{100}{ln~100} = 21.7$ $f(1000) = \frac{1000}{ln~1000} = 144.8$ $f(10,000) = \frac{10,000}{ln~10,000} = 1085.7$ $f(100,000) = \frac{100,000}{ln~100,000} = 8685.9$ $f(1,000,000) = \frac{1,000,000}{ln~1,000,000} = 72,382$ Since the values seem to increase without bound, we could guess that: $\lim\limits_{x \to \infty}f(x) = \infty$ (c) We can see a sketch of the graph below.
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