anonymous
  • anonymous
The limit as x approaches infinity of the integral from x to 2x of dt/t
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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abb0t
  • 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
  • anonymous
\[\lim_{x \rightarrow \infty}\int\limits_{x}^{2x}\frac{ dt }{ t }\]
anonymous
  • anonymous
Do you know the answer?

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More answers

anonymous
  • anonymous
Because I believe the integral simply diverges.
anonymous
  • anonymous
Like the limit doesn't exist.
anonymous
  • anonymous
the answer is just ln2, but I don't know how my teacher got that
anonymous
  • anonymous
Hmm... One moment.
anonymous
  • anonymous
Okay
anonymous
  • anonymous
I am seriously stuck. I would love to know how yo approach this.
anonymous
  • anonymous
I multiplied the top and bottom by x and applied L'hospitals rule but that doesn't help.
anonymous
  • anonymous
I'm stuck too, I've been stuck on this problem for a while now. Thanks anyways :)
anonymous
  • anonymous
Well zzr0ck3r is replying. He is pretty good.
zzr0ck3r
  • zzr0ck3r
nah just typing so i can see:)
zzr0ck3r
  • zzr0ck3r
im stuck
anonymous
  • anonymous
X) .
anonymous
  • anonymous
I really hope he can help me out , I have a test tomorrow and I need to figure this out
anonymous
  • anonymous
Well I tried to multiply the top and bottom by x in order to apply l'hospitals rule.
anonymous
  • anonymous
But the problem is that we still get infinity
anonymous
  • anonymous
So I can't really do anything. Applying L'hospital's rule again doesn't help either.
anonymous
  • anonymous
But can't l'hopital's rule be applied more than once ?
anonymous
  • anonymous
Yeah I know. But that makes the situation worse.
anonymous
  • anonymous
Yeah that's true...
zzr0ck3r
  • zzr0ck3r
ok got it
anonymous
  • anonymous
Does it involve splitting up the integral? Then I think I got it too.
zzr0ck3r
  • 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
  • zzr0ck3r
im not sure if you can do that but it looks good to me:)
anonymous
  • anonymous
ln(0) is undefined.
zzr0ck3r
  • zzr0ck3r
ahh we could have taken the integral to 1
anonymous
  • anonymous
Pick some other value like 2 :P .
zzr0ck3r
  • zzr0ck3r
yeah...
zzr0ck3r
  • zzr0ck3r
ok have fun, tv show with the wife:)
anonymous
  • anonymous
What about the limit though?
anonymous
  • anonymous
But okay have fun X) .
anonymous
  • anonymous
NVM I got the limit part too.
anonymous
  • anonymous
@jossy04 : did you get what he did?
anonymous
  • anonymous
So the way zzr0ck3r did it is wrong ??
anonymous
  • anonymous
No it's right haha,
anonymous
  • anonymous
Yeah I get what he did
anonymous
  • anonymous
But when he split up the intergal he used the wrong limits.
anonymous
  • anonymous
Okay haha thanks :)
anonymous
  • anonymous
Because ln(0) is undefined at 0.
anonymous
  • anonymous
He should have split it up at like 1 or 2 or something non negative or 0.
anonymous
  • anonymous
Ohh okay then .
anonymous
  • 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
  • anonymous
Yep I can do that.
anonymous
  • anonymous
Can you help me please ? :)
anonymous
  • anonymous
First notice that if that thing is actually approaching 1 note how the function become 0/0 right?
anonymous
  • anonymous
Because the integral doesn't move anywhere it's just going to be a 0.
anonymous
  • anonymous
Does that make sense?
anonymous
  • anonymous
Yeah it makes sense .
anonymous
  • anonymous
Alright we know l'hospitals rule can be applied if the limit approaches 0/0 or infinity/infinity
anonymous
  • anonymous
Do you know the fundamental theorem of Calculus?
tkhunny
  • 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
  • anonymous
I can't believe I didn't hink of Logarithmic properties. I was stupid :( .
anonymous
  • 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
  • anonymous
Do you understand that so far?
anonymous
  • anonymous
What tkhunny did was so much easier haha Yeah I'm understanding . Thanks @tkhunny :)
anonymous
  • anonymous
Allright. Do you get how the fundamental theorem of Calculus works?
anonymous
  • anonymous
Yeah I get how it works .
anonymous
  • 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
  • anonymous
The derivative of the bottom is just 2x
anonymous
  • anonymous
So we get our limit as: |dw:1377754905563:dw|
anonymous
  • anonymous
I know you can do that :P .
anonymous
  • anonymous
Do you understand?
anonymous
  • anonymous
Yeah I get it
anonymous
  • anonymous
:) .
anonymous
  • anonymous
so it would end up being cos(1)/2 ???
anonymous
  • anonymous
yeah!
anonymous
  • anonymous
Okay then thank very much for your help :D

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