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: 55

Answer

The solution is: $$\lim_{x\to0^+}x^{2x}=1.$$

Work Step by Step

To solve this limit we will first transform the expression under the limit. We will use the fact that the natural exponential function and the natural logarithmic function are inverse to each other so $a=e^{\ln a}$. This means that $x^{2x}=e^{\ln x^{2x}}=e^{2x\ln x}$. We additionally used the logarithm rule $\ln b^a=a\ln b$. Our limit becomes $$\lim_{x\to0^+}x^{2x}=\lim_{x\to0^+}e^{2x\ln x}.$$ Now we can use the fact that the exponential function is continuous which allows us to exchange its place with the limit: $$\lim_{x\to0^+}x^{2x}=\lim_{x\to0^+}e^{2x\ln x}=e^{\lim_{x\to0^+}2x\ln x}=e^{2\lim_{x\to0^+}x\ln x}=e^{2l},$$ where we have denoted $l=\lim_{x\to0^+}x\ln x$. Now we have to solve this limit. We will do so by using L'Hopital's rule. "LR" will denote "Apply L'Hopital's rule". $$l=\lim_{x\to0^+}x\ln x=\lim_{x\to0^+}\frac{\ln x}{\frac{1}{x}}=\left[\frac{\ln0^+}{\frac{1}{0^+}}\right]=\left[\frac{-\infty}{\infty}\right][\text{LR}]=\lim_{x\to0^+}\frac{(\ln x)'}{\left(\frac{1}{x}\right)'}=\lim_{x\to0^+}\frac{\frac{1}{x}}{-\frac{1}{x^2}}=\lim_{x\to0^+}\frac{-x^2}{x}=\lim_{x\to0^+}-x=-0=0.$$ Putting this back into the original limit we get $$\lim_{x\to0^+}x^{2x}=e^{2l}=e^{2\cdot0}=1.$$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.