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

Status
Not open for further replies.

#### Chris L T521

##### Well-known member
Staff member
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:
Use L'Hôpital's rule.

#### Chris L T521

##### Well-known member
Staff member
This week's problem was correctly answered by MarkFL and Pranav. You can find Mark's solution below.

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:

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

Status
Not open for further replies.