Problem of the Week #87 - November 25th, 2013

[Chris L T521]

Thanks again to those who participated in last week's POTW! Here's this week's problem!

-----

Problem: Evaluate $\displaystyle \lim_{x\to 0}\frac{1}{x}\int_0^x \left(1-\tan(2t)\right)^{1/t}\,dt$.

-----

Hint: [sp]Use L'Hôpital's rule.[/sp]

 

[Chris L T521]

This week's problem was correctly answered by MarkFL and Pranav. You can find Mark's solution below.

[sp]We are given to evaluate:

\(\displaystyle L=\lim_{x\to 0}\left(\frac{1}{x}\int_0^x \left(1-\tan(2t)\right)^{1/t}\,dt \right)\)

Since we have the indeterminate form \(\displaystyle \frac{0}{0}\), application of L'Hôpital's rule yields:

\(\displaystyle L=\lim_{x\to 0}\left(\left(1-\tan(2x)\right)^{1/x} \right)\)

Taking the natural log of both sides (and applying the rules of logs as they apply to limits and exponents), we obtain:

\(\displaystyle \ln(L)=\lim_{x\to 0}\left(\frac{\ln\left(1-\tan(2x) \right)}{x} \right)\)

Since we have the indeterminate form \(\displaystyle \frac{0}{0}\), application of L'Hôpital's rule yields:

\(\displaystyle \ln(L)=2\lim_{x\to 0}\left(\frac{\sec^2(2x)}{\tan(2x)-1} \right)=-2\)

Converting from logarithmic to exponential form, we find:

\(\displaystyle L=e^{-2}\)

Hence, we conclude:

\(\displaystyle \lim_{x\to 0}\left(\frac{1}{x}\int_0^x \left(1-\tan(2t)\right)^{1/t}\,dt \right)=\frac{1}{e^2}\)[/sp]