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: 37


$$\lim_{x\to0^+}\frac{\arctan(2x)}{\ln x}=0$$

Work Step by Step

$$A=\lim_{x\to0^+}\frac{\arctan(2x)}{\ln x}$$ As $x$ approaches $0^+$, $\arctan(2x)$ approaches $\arctan(2\times 0)=\arctan0=0$ However, at the same time, $\ln x$ approaches $\ln0^+$, meaning it approaches $-\infty$. So there is no indeterminate form here, and L'Hospital's Rule cannot be applied. Nevertheless, we see that while the denominator approaches a madly big number, the numerator instead approaches $0$. That means the whole formula still reaches $0$ in the end, for $\frac{0}{a}=0$ ($a$ stands for any real number) In other words, $$A=0$$
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