Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 4 - Section 4.4 - Indeterminate Forms and l''Hospital''s Rule - 4.4 Exercises - Page 312: 55


$$\lim_{x\to\infty}(x-\ln x)=\infty$$

Work Step by Step

$$A=\lim_{x\to\infty}(x-\ln x)$$ We need to do some modifications here, which would change $x$ into a logarithmic function somehow. We see that $\ln e^x=x$. So we can rewrite $x$ with $\ln e^x$. $$A=\lim_{x\to\infty}(\ln e^x-\ln x)$$ $$A=\lim_{x\to\infty}\ln\Big(\frac{e^x}{x}\Big)$$ $$A=\ln\Big(\lim_{x\to\infty}\frac{e^x}{x}\Big)$$ $$A=\ln B$$ We see that as $x\to\infty$, $e^x\to\infty$. So we have an indeterminate form of $\infty/\infty$, applicable to L'Hospital's Rule: $$B=\lim_{x\to\infty}\frac{(e^x)'}{x'}$$ $$B=\lim_{x\to\infty}\frac{e^x}{1}$$ $$B=\lim_{x\to\infty}e^x$$ $$B=\infty$$ Therefore, $$A=\ln B=\ln(\infty)=\infty$$
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