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: 38


$$\lim_{x\to0^+}\frac{x^x-1}{\ln x+x-1}=0$$

Work Step by Step

$$A=\lim_{x\to0^+}\frac{x^x-1}{\ln x+x-1}$$ 1) The problem in this exercise is to calculate $\lim_{x\to0^+}(x^x)$. Let's call this B. $$B=\lim_{x\to0^+}(x^x)$$ In order to deal with $B$, we need to do a little transformation first. Take $x^x=(e^{\ln x})^x=e^{x\ln x}$ (for $e^{\ln x}=x$) So we have, $$B=\lim_{x\to0^+}(e^{x\ln x})$$ $$B=e^{\lim_{x\to0^+}(x\ln x)}$$ We now call $$C=\lim_{x\to0^+}x\ln x$$ Divide both numerator and denominator by $x$: $$C=\lim_{x\to0^+}\frac{\ln x}{\frac{1}{x}}$$ As $x\to0^+$, $(\ln x)\to-\infty$ and $\frac{1}{x}\to\infty$. An indeterminate form of $-\infty/\infty$, so we apply L'Hospital's Rule: $$C=\lim_{x\to0^+}\frac{\frac{1}{x}}{-\frac{1}{x^2}}$$ $$C=\lim_{x\to0^+}-\frac{x^2}{x}$$ $$C=\lim_{x\to0^+}(-x)$$ $$C=-0=0$$ Consequently, $$B=e^{\lim_{x\to0^+}(x\ln x)}=e^0=1$$ That means, $$\lim_{x\to0^+}(x^x)=1$$ We now come back to the main question of $A$. 2) $\lim_{x\to0^+}(x^x-1)=1-1=0$ and $\lim_{x\to0^+}(\ln x+x-1)=-\infty$ (As $x\to0^+$, $\ln x+x-1\to(-\infty+0-1)=-\infty$) That takes us to the form of $\frac{0}{\infty}$, which is no different from $0$. Therefore, $$A=0$$
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