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: 92

Answer

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

Work Step by Step

To solve this limit we will use the fact that the natural exponential and the natural logarithmic functions are inverse to each other i.e. that $e^{\ln a}=a.$ This yields $$x^{1/\ln x}=e^{\ln x^{1/\ln x}}=e^{\frac{1}{\ln x}\ln x}=e^1=e.$$ Now, putting this into the limit we get $$\lim_{x\to0^+}x^{1/\ln x}=\lim_{x\to0^+}e=e.$$
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