# Lim of integral

1. Oct 18, 2004

### 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

Evaluating it with maple gives $$e^{-2}$$

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

Last edited by a moderator: Oct 18, 2004
2. Oct 18, 2004

### 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.

3. Oct 18, 2004

### tangur

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

4. Oct 18, 2004

### 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

5. Oct 19, 2004

### 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.

6. Oct 19, 2004

### arildno

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

7. Oct 19, 2004

### NateTG

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

8. Oct 19, 2004

### jatin9_99

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