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Topic: Limits (Read 1205 times) |
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Braincramps
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Let y = x1/ [smiley=surd.gif](1-x). Find the limit of y/(1-y) as x [smiley=longrightarrow.gif] 1-.
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ThudnBlunder
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Re: Limits
« Reply #1 on: May 16th, 2004, 1:38pm » |
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Have you attempted it?
If so, how far did you get?
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Benoit_Mandelbrot
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Re: Limits
« Reply #2 on: May 17th, 2004, 9:21am » |
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The answer is :: infinity. Here's why. We need to substitute into the fraction so the limit can be evaluated. Because 1-x is greater than zero when x goes to one from the left, the square root of 1-1 is zero, and because that limit would be positive, then 1/sqrt(1-x) approaches infinity. 1^infinity is one. In the denominator, we get a negative number because x goes to one from the left. So we factor the negative into the denominator get 1/(1-u) as u goes to 1 from the left. The limit is infinity. ::
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| « Last Edit: May 19th, 2004, 9:06am by Benoit_Mandelbrot » |
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Icarus
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Re: Limits
« Reply #3 on: May 17th, 2004, 6:17pm » |
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1[supinfty] need not be 1. For instance, limx[to]0 (ex)1/x = e
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TenaliRaman
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QED!! 
(put the graph into words and using the definition of limits, i think we can finish it off quite easily!)
(The white line is y and the red is y/1-y)
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| « Last Edit: May 17th, 2004, 6:52pm by TenaliRaman » |
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Benoit_Mandelbrot
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Re: Limits
« Reply #5 on: May 18th, 2004, 6:03am » |
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True, Icarus, but in this case, the x-es don't cancel out, so you can't do direct substitution. In the case of the problem, 1^[infty] would be one.
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Icarus
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Re: Limits
« Reply #6 on: May 18th, 2004, 10:05pm » |
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on May 18th, 2004, 6:03am, Benoit_Mandelbrot wrote:| In the case of the problem, 1^[infty] would be one. |
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Perhaps, but you have to prove it.
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"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
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ThudnBlunder
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Re: Limits
« Reply #7 on: May 19th, 2004, 4:37am » |
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on May 17th, 2004, 6:50pm, TenaliRaman wrote:QED!!
(put the graph into words and using the definition of limits, i think we can finish it off quite easily!)
(The white line is y and the red is y/1-y) |
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Oh, you use Graphmatica, too!
Try LIM Sin(1/x)
x->0
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Barukh
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Re: Limits
« Reply #8 on: May 19th, 2004, 5:39am » |
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What’s wrong with this?
Substitute: x = 1 – z, and we seek the limit when z [to] 0+.
ln(y) = ln[ (1-z) 1/[sqrt]z ] = z-1/2 ln(1-z) = z-1/2 ( -z - z2/2 - z3/3 - ...) = - (z1/2 + z3/2/2 + z5/2/3 + ...) [to] 0 when z [to] 0+.
Therefore y [to] 1-. 
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| « Last Edit: May 19th, 2004, 5:41am by Barukh » |
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Benoit_Mandelbrot
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Re: Limits
« Reply #9 on: May 19th, 2004, 6:04am » |
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Nice .gif Barukh! How did you make that or get that? How would I make something like that, like a supernova in the correct size?
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towr
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Re: Limits
« Reply #10 on: May 19th, 2004, 6:05am » |
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on May 19th, 2004, 5:39am, Barukh wrote:| Therefore y [to] 1-. |
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We're looking for the limit for y/(1-y)
so that would go to [infty] then..
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Wikipedia, Google, Mathworld, Integer sequence DB
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Barukh
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Re: Limits
« Reply #11 on: May 19th, 2004, 7:42am » |
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on May 19th, 2004, 6:04am, Benoit_Mandelbrot wrote:| Nice .gif Barukh! How did you make that or get that? How would I make something like that, like a supernova in the correct size? |
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Yes, it's nice, but it’s not mine – you should ask TenaliRaman
on May 19th, 2004, 6:05am, towr wrote:We're looking for the limit for y/(1-y)
so that would go to [infty] then.. |
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So, what's the flaw in my derivation? Or, are you saying that's not enough to establish the value of y/(1-y)?
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towr
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Re: Limits
« Reply #12 on: May 19th, 2004, 8:53am » |
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on May 19th, 2004, 7:42am, Barukh wrote:| So, what's the flaw in my derivation? Or, are you saying that's not enough to establish the value of y/(1-y)? |
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I don't see why there would be any flaw. It adequately shows y/(1-y) is [infty] in the limit, just as TenaliRaman's graph shows
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| « Last Edit: May 19th, 2004, 8:54am by towr » |
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TenaliRaman
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Re: Limits
« Reply #13 on: May 21st, 2004, 10:45am » |
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on May 19th, 2004, 4:37am, THUDandBLUNDER wrote:
Oh, you use Graphmatica, too!
Try LIM Sin(1/x)
x->0 |
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finite value <= |1|
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ThudnBlunder
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Re: Limits
« Reply #14 on: May 23rd, 2004, 1:10pm » |
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What is LIM[smiley=surd.gif][x(x + 1)] - x?
x -> oo
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| « Last Edit: May 23rd, 2004, 2:36pm by ThudnBlunder » |
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Barukh
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Re: Limits
« Reply #15 on: May 24th, 2004, 12:06am » |
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on May 23rd, 2004, 1:10pm, THUDandBLUNDER wrote:What is LIM[smiley=surd.gif][x(x + 1)] - x?
x -> oo |
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[smiley=blacksquare.gif]
1/2.
Proof: Let a = [sqrt] [x(x+1)], b = x + 1/2. Then: (b-a)(b+a) = b2 – a2 = 1/4, therefore b – a [to] 0 when x [to] [infty].
[smiley=blacksquare.gif]
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TenaliRaman
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Re: Limits
« Reply #16 on: May 24th, 2004, 11:32am » |
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wowee,
that's elegant barukh.
much pretty straightforward method that i usually use,
::multiply numerator and denominator by sqrt(x(x+1))+x.note that this is well within the rules since x->oo and hence x!=oo.then divide numerator by x.again well within the rules.then just take x->oo and et voila 1/2::
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Barukh
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Re: Limits
« Reply #17 on: May 24th, 2004, 11:08pm » |
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on May 24th, 2004, 11:32am, TenaliRaman wrote:wowee,
that's elegant barukh. |
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Thanks.
Quote:| much pretty straightforward method that i usually use, |
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I'm having difficulty to understand your method. The first operation seems OK, but after the second one - dividing numerator by x - you are dealing with another limit.
Or did you mean: dividing both numerator and denominator by x?
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Sir Col
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Re: Limits
« Reply #18 on: May 25th, 2004, 2:32am » |
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I think that's what he meant. In which case...
L = x/([sqrt][x(x+1)]+x) = 1/([sqrt][(x+1)/x]+1)
As x[to][infty], (x+1)/x[to]1, so L[to]1/2.
But as TenaliRaman said, what an elegant method, Barukh! *round of applause smilie*
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Barukh
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Re: Limits
« Reply #19 on: May 25th, 2004, 4:25am » |
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on May 25th, 2004, 2:32am, Sir Col wrote:I think that's what he meant. In which case...
L = x/([sqrt][x(x+1)]+x) = 1/([sqrt][(x+1)/x]+1)
As x[to][infty], (x+1)/x[to]1, so L[to]1/2. |
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Really nice! 
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ThudnBlunder
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Re: Limits
« Reply #20 on: May 31st, 2004, 7:05pm » |
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If f(x,y)= x4logexy find LIM f(x,y) as (x,y)->(0,0)
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| « Last Edit: May 31st, 2004, 7:06pm by ThudnBlunder » |
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Eigenray
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Re: Limits
« Reply #21 on: May 31st, 2004, 9:32pm » |
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on May 31st, 2004, 7:05pm, THUDandBLUNDER wrote:| If f(x,y)= x4logexy find LIM f(x,y) as (x,y)->(0,0) |
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Try letting (x,y)->(0,0) along the curve y=e^(-1/x^n) for n=3,4,5.
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