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Bacchus
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Registered: 10/10/06
Posts: 914
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Rusty on my calculus.
#15168976 - 10/02/11 08:37 PM (12 years, 7 months ago) |
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I'm being a hypocrite in the fact that I get annoyed when I see homework questions here, but now I'm going to post one because I'm taking calculus III after a long time off from math. I need to determine if the sequence converges or diverges as n approaches infinity. If it converges, what is the limit?
I can reason that as n goes to infinity, the numerator will swing between -1 and 1 and the denominator will increase without bound, so the expression converges to zero.
How do I formalize the logic, though? sin(2n) has no limit as n goes to infinity, so I can't take the limit of the top and bottom, can I? L'Hosptital's rule doesn't apply. Where do I go? Is it enough to say ?
A second problem that I'm stuck on is  I multiply by to get I've played with it from here but haven't made any progress. I've used L'Hospital's rule twice, pulled out the constant and rewrote the fraction as 
From here, I can keep applying L'Hospital's rule, but it doesn't get me anywhere. In the end, I'm still left with either infinity times infinity or infinity times zero indeterminate forms.
I tried putting this one into Wolfram|Alpha so I could expand it out to see the steps, but all I got was this:
Edited by Bacchus (10/02/11 09:55 PM)
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johnm214


Registered: 05/31/07
Posts: 17,582
Loc: Americas
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Re: Rusty on my calculus. [Re: Bacchus]
#15170466 - 10/03/11 06:51 AM (12 years, 7 months ago) |
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The first case would converge on 0, as you say.
The numerator will vacillate between positive and negative numbers including magnitude 1 through 0, and the denominator gets large, so approaching infinity you have a finite number over an essentially infinite number which aproaches 0.
You can show this to be the case a number of ways, but basically if your function has f1(x)/f2(x) if the degree of f1 is less than f2 it will approach 0 at infinity. If your prof covered some other way you can use that way, but this kind of problem is intuitive.
Fort the second one, you can use the same rule: The numerator has degree of 1 and the denominator has degree of 2, so it goes to 0 at infinity.
Intuitively, you know 1 to any power is 1, so at infinity the magnitude of the numerator is n and the denominator is nn.
Been a while since I've done this stuff as well, but seems pretty straight forward to me. Check out wikipedia for more discussion if you want: http://en.wikipedia.org/wiki/Limit_of_a_function#Properties
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DieCommie

Registered: 12/11/03
Posts: 29,258
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Re: Rusty on my calculus. [Re: Bacchus]
#15171492 - 10/03/11 12:07 PM (12 years, 7 months ago) |
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Quote:
From here, I can keep applying L'Hospital's rule, but it doesn't get me anywhere.
You sure about that? Looks like you can apply it a couple more times to me.
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indraclee
friend



Registered: 09/28/11
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Re: Rusty on my calculus. [Re: Bacchus]
#15172569 - 10/03/11 04:27 PM (12 years, 7 months ago) |
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All I think I got these figured out. Tell me if you agree cuz been a year or two since I've done this kinda stuff.
First off I think that both of these converge. The first I used the limit comparison test and the second I used the alternating series test.
1) You know that sin(2n) is always less than or equal to one. Therefore you can say that |a_n| <= 1/(1+n^2).
Now because 1/(1+n^2) definitely converges as n->inf and because a_n is always less than or equal to 1/(1+n^2), a_n must also converge 
2) If you write out a few terms of this one (0,1/2,-2/5,3/10), you can see it is monotone decreasing. Taking the absolute value you are left with n/(n^2+1) for a_n. This definitely converges as n->inf. These are the only requirements for the alternating series test.
Tell me if this helps. I can try to be more descriptive.
-------------------- Wake up to find out that you are the eyes of the world.
 
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Bacchus
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Registered: 10/10/06
Posts: 914
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Re: Rusty on my calculus. [Re: indraclee]
#15173827 - 10/03/11 08:12 PM (12 years, 7 months ago) |
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Quote:
DieCommie said:
Quote:
From here, I can keep applying L'Hospital's rule, but it doesn't get me anywhere.
You sure about that? Looks like you can apply it a couple more times to me.
Yeah, that's why I rewrote the fraction with n^3 in the numerator. I wanted to bring it down to 1 so that I no longer had an indeterminate form, but you end up with Which is nonreal. So apply L'Hospital's rule one more time to bring the denominator back to (-1)^(-n), and I'm right back at 0*infinity.
indraclee, you totally nailed it on both of them. I was scratching my head at how to actually use the squeeze and absolute value theorems. Your explanation makes perfect sense. Thanks.
Edited by Bacchus (10/03/11 08:18 PM)
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Bacchus
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Registered: 10/10/06
Posts: 914
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Re: Rusty on my calculus. [Re: DieCommie]
#15173839 - 10/03/11 08:15 PM (12 years, 7 months ago) |
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deletemepleasethanks
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Living on a no-Flash diet is way easier than you think. Give it a shot.
Edited by Bacchus (10/03/11 08:17 PM)
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