1. ## really hard limit of an integral!

limit as x goes to 0 of (1/x) * the integral from 0 to x of (1-tan(2t))^(1/t) dt

2. Originally Posted by stones44
limit as x goes to 0 of (1/x) * the integral from 0 to x of (1-tan(2t))^(1/t) dt
Is this what it looks like?

$\displaystyle \lim_{x->0}\frac{1}{x}\int_{x}^0(1-tan(2t))^{\frac{1}{t}}dt$

If so, I will help you work on it. I haven't worked it out yet, but I think that we need to use the squeeze theorem. Also remember that the function $\displaystyle \int_{x}^0(1-tan(2t))^{\frac{1}{t}}dt=-\int_{0}^x(1-tan(2t))^{\frac{1}{t}}dt$

3. flip the limits and yes

so

$\displaystyle \lim_{x->0}\frac{1}{x}\int_{0}^x(1-tan(2t))^{\frac{1}{t}}dt$

4. i dont think i know that theorem yet so i doubt that is the way i am expected to solve it

5. Originally Posted by stones44
limit as x goes to 0 of (1/x) * the integral from 0 to x of (1-tan(2t))^(1/t) dt
$\displaystyle \lim_{x \rightarrow 0} \frac{\int_0^x (1 - \tan(2t))^{1/t} \, dt}{x} = \lim_{x \rightarrow 0} (1 - \tan(2x))^{1/x}$

using l'Hopital's Rule and the Fundamental Theorem of Calculus.

Now note that $\displaystyle (1 - \tan(2x))^{1/x} = e^{\ln (1 - \tan(2x))^{1/x}} = e^{\frac{1}{x} \ln (1 - \tan(2x))}$

so you should now use l'Hopital's rule to find $\displaystyle \lim_{x \rightarrow 0} \frac{\ln (1 - \tan(2x))}{x}$.