lim of integral

[tangur]

\lim_{x \rightarrow 0} \frac{1}{x} \int_0^x {(1-tan2t)}^{\frac{1}{t}} dt

and nevermind the { in front of the 0, i couldnt figure out how to take it out, its my first time posting

I can't figure out how to start attacking this problem, do I have to intregrate by parts? if so what do I use as u and v.
I tried thinking of any identities that could help but nothing came to mind.

Any help will be greatly appreciated

Thanks in advance.

Evaluating it with maple gives e^{-2}

Admin note: pesky { removed. Click the latex image to see its code.

[NateTG]

Well, you could re-write it as:
\lim_{x\rightarrow 0} \frac{ \int_0^x {(1-tan2t)}^{\frac{1}{t}} dt}{x}
And then apply l'Hopital's rule.

No clue if that's the right way to go, and I'm too lazy to check.

[tangur]

you would apply lhopital, take the limit and then diifferentiate?

[tangur]

oh damn just as i was writing this message i figured out how to apply lhopital , so simple but it escaped by grasp, its just a differentiation of an integral, thx so much

[NateTG]

\lim_{x\rightarrow 0} \frac{ \int_0^x {(1-tan2t)}^{\frac{1}{t}} dt}{x}
Take the derivatives of the top and bottom:
\lim_{x \rightarrow 0} \frac{(1-tan(2x))^{\frac{1}{x}}}{1}
\lim_{x \rightarrow 0} (1 - \tan (2x))^{\frac{1}{x}}

But now, the expression inside the parens goes to 1 since \tan(0)=0 and the exponent should also tend to bring things to 1, so the limit should be 1.

[arildno]

This is wrong, Nate G!
You've got the indeterminate form 1^{\infty}
we have:
\lim_{x\to0}(1-tan(2x))^{\frac{1}{x}}=e^{\lim_{x\to0}\frac{ln(1-tan(2x))}{x}}
=e^{\lim_{x\to0}\frac{-2x}{x}}=e^{-2}

[NateTG]

This is wrong, Nate G!
You've got the indeterminate form 1^{\infty}
we have:
\lim_{x\to0}(1-tan(2x))^{\frac{1}{x}}=e^{\lim_{x\to0}\frac{ln(1-tan(2x))}{x}}
=e^{\lim_{x\to0}\frac{-2x}{x}}=e^{-2}

Stupid brain....for some strange reason I thought it was going to 1^0. <Puts on pointy hat with donkey ears.>

[jatin9_99]

just take the integral first then it will be real easy to do this question