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!

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

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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:

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

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

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

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

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

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

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

Converting from logarithmic to exponential form, we find:

$$L=e^{-2}$$

Hence, we conclude:

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