The limit as x approaches infinity of the integral from x to 2x of dt/t

- anonymous

The limit as x approaches infinity of the integral from x to 2x of dt/t

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- katieb

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- abb0t

What? I think I know what you're talking about, and I htink you're going to use L'hopitals rule, but can you re-write it using LaTex?

- anonymous

\[\lim_{x \rightarrow \infty}\int\limits_{x}^{2x}\frac{ dt }{ t }\]

- anonymous

Do you know the answer?

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## More answers

- anonymous

Because I believe the integral simply diverges.

- anonymous

Like the limit doesn't exist.

- anonymous

the answer is just ln2, but I don't know how my teacher got that

- anonymous

Hmm... One moment.

- anonymous

Okay

- anonymous

I am seriously stuck. I would love to know how yo approach this.

- anonymous

I multiplied the top and bottom by x and applied L'hospitals rule but that doesn't help.

- anonymous

I'm stuck too, I've been stuck on this problem for a while now. Thanks anyways :)

- anonymous

Well zzr0ck3r is replying. He is pretty good.

- zzr0ck3r

nah just typing so i can see:)

- zzr0ck3r

im stuck

- anonymous

X) .

- anonymous

I really hope he can help me out , I have a test tomorrow and I need to figure this out

- anonymous

Well I tried to multiply the top and bottom by x in order to apply l'hospitals rule.

- anonymous

But the problem is that we still get infinity

- anonymous

So I can't really do anything. Applying L'hospital's rule again doesn't help either.

- anonymous

But can't l'hopital's rule be applied more than once ?

- anonymous

Yeah I know. But that makes the situation worse.

- anonymous

Yeah that's true...

- zzr0ck3r

ok got it

- anonymous

Does it involve splitting up the integral? Then I think I got it too.

- zzr0ck3r

\[\int_{x}^{2x}\frac{1}{t}dt=-\int_{0}^{x}\frac{1}{t}dt+\int_{0}^{2x}\frac{1}{t}dt=\\-\ln(x)+\ln(0)+\ln(2x)-\ln(0)=-\ln(x)+\ln(2)+\ln(x) = \ln(2)\]

- zzr0ck3r

im not sure if you can do that but it looks good to me:)

- anonymous

ln(0) is undefined.

- zzr0ck3r

ahh we could have taken the integral to 1

- anonymous

Pick some other value like 2 :P .

- zzr0ck3r

yeah...

- zzr0ck3r

ok have fun, tv show with the wife:)

- anonymous

What about the limit though?

- anonymous

But okay have fun X) .

- anonymous

NVM I got the limit part too.

- anonymous

@jossy04 : did you get what he did?

- anonymous

So the way zzr0ck3r did it is wrong ??

- anonymous

No it's right haha,

- anonymous

Yeah I get what he did

- anonymous

But when he split up the intergal he used the wrong limits.

- anonymous

Okay haha thanks :)

- anonymous

Because ln(0) is undefined at 0.

- anonymous

He should have split it up at like 1 or 2 or something non negative or 0.

- anonymous

Ohh okay then .

- anonymous

@Dido525 do you happen to know how to solve this one too ? \[\lim_{x \rightarrow 1}\frac{ \int\limits_{1}^{x}\cos tdt }{ x^{2}-1 }\]

- anonymous

Yep I can do that.

- anonymous

Can you help me please ? :)

- anonymous

First notice that if that thing is actually approaching 1 note how the function become 0/0 right?

- anonymous

Because the integral doesn't move anywhere it's just going to be a 0.

- anonymous

Does that make sense?

- anonymous

Yeah it makes sense .

- anonymous

Alright we know l'hospitals rule can be applied if the limit approaches 0/0 or infinity/infinity

- anonymous

Do you know the fundamental theorem of Calculus?

- tkhunny

People! What are you doing? Just evaluate it directly. No l'Hopital, no scary limits, no splitting up.
For some x > e (really, just pick a positive number. I used 'e' just for fun. Convergence doesn't care about some small finite area at the beginning.)
\(\int\limits_{x}^{2x}\dfrac{1}{t}\;dt = ln(t)|_{x}^{2x} = ln(2x) - ln(x) = ln\left(\dfrac{2x}{x}\right) = ln(2)\)
The result is independent of x and the limit is of no consequence.

- anonymous

I can't believe I didn't hink of Logarithmic properties. I was stupid :( .

- anonymous

So L'hospitals rule says if we get a limit in the form 0/0 we take the limits of the derivatives in the numerator and denominator.

- anonymous

Do you understand that so far?

- anonymous

What tkhunny did was so much easier haha
Yeah I'm understanding .
Thanks @tkhunny :)

- anonymous

Allright. Do you get how the fundamental theorem of Calculus works?

- anonymous

Yeah I get how it works .

- anonymous

Alright, so we have that x in that integral. We substitute x instead of a t inside that cosine and multiply by the derivative of x which is just 1.

- anonymous

The derivative of the bottom is just 2x

- anonymous

So we get our limit as:
|dw:1377754905563:dw|

- anonymous

I know you can do that :P .

- anonymous

Do you understand?

- anonymous

Yeah I get it

- anonymous

:) .

- anonymous

so it would end up being cos(1)/2 ???

- anonymous

yeah!

- anonymous

Okay then thank very much for your help :D

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