Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 6 - Exponential, Logarithmic, And Inverse Trigonometric Functions - 6.5 L'Hopital's Rule; Indeterminate Forms - Exercises Set 6.5 - Page 448: 35

Answer

$+\infty$

Work Step by Step

Our aim is to evaluate the limit for $\lim\limits_{x \to +\infty} [x-\ln (x^2+1)]$ Let us consider that $y=x-\ln (x^2+1) \implies e^y=\dfrac{e^x}{x^2+1}$ But $\lim\limits_{x \to +\infty} \dfrac{e^x}{x^2+1}=\dfrac{\infty}{\infty}$ We can see that the numerator and denominator have a limit of $\infty$, so the limit shows the indeterminate form of type $\dfrac{\infty}{\infty}$. So, we will apply L'Hopital's rule which can be defined as: $\lim\limits_{x \to a} \dfrac{P(x) }{Q(x)}=\lim\limits_{x \to a}\dfrac{P'(x)}{Q'(x)}$ where, $a$ can be any real number, infinity or negative infinity. $\lim\limits_{x \to +\infty} \dfrac{e^x}{x^2+1}=\lim\limits_{x \to +\infty} \dfrac{e^x}{2x}=+ \infty$ Thus, we have $\lim\limits_{x \to +\infty} y=\ln(+\infty)=+\infty$
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