Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 4 - Applications of the Derivative - 4.7 L'Hopital's Rule - 4.7 Exercises: 99


The solution is $$\lim_{x\to0^+}x^{1/(1+\ln x)}=e$$

Work Step by Step

To solve this limit we will use the fact that the natural exponential and the natural logarithmic function are inverse to each other i.e. $e^{\ln a}=a$. Using this we have $x^{1/(1+\ln x)}=e^{\ln x^{1/(1+\ln x)}}=e^{\frac{1}{1+\ln x}\ln x}=e^{\frac{\ln x}{1+\ln x}}$. Now putting this into the limit and using the fact that the natural exponential function is continuous (so it can exchange places with the limit) we have: $$L=\lim_{x\to0^+}x^{1/(1+\ln x)}=\lim_{x\to0^+}e^{\frac{\ln x}{1+\ln x}}=e^{\lim_{x\to0^+}\frac{\ln x}{1+\ln x}}=e^l,$$ where we have denoted $l=\lim_{x\to0^+}\frac{\ln x}{1+\ln x}$. Now let us solve this limit. $$l=\lim_{x\to0^+}\frac{\ln x}{1+\ln x}=\lim_{x\to0^+}\frac{\ln x}{\ln x\left(\frac{1}{\ln x}+1\right)}=\lim_{x\to0^+}\frac{1}{\frac{1}{\ln x}+1}=\left[\frac{1}{\frac{1}{\ln0^+}+1}\right]=\left[\frac{1}{\frac{1}{-\infty}+1}\right]=\left[\frac{1}{0+1}\right]=1.$$ Returning this into the initial limit we have $$L=e^l=e^1=e.$$
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