wu :: forums
        (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi)
      

        riddles >> putnam exam (pure math) >> Limits
        
(Message started by: Braincramps on May 16th, 2004, 12:25pm)
Title: Limits
Post by Braincramps on May 16th, 2004, 12:25pm
Let y = x1/ [smiley=surd.gif](1-x).  Find the limit of y/(1-y) as x [smiley=longrightarrow.gif] 1-.
Title: Re: Limits
Post by THUDandBLUNDER on May 16th, 2004, 1:38pm
Have you attempted it?
If so, how far did you get?
:-X
Title: Re: Limits
Post by Benoit_Mandelbrot on May 17th, 2004, 9:21am
The answer is :: [hide]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. [/hide]::
Title: Re: Limits
Post by Icarus on May 17th, 2004, 6:17pm
1[supinfty] need not be 1. For instance, limx[to]0 (ex)1/x = e
Title: Re: Limits
Post by TenaliRaman on May 17th, 2004, 6:50pm
QED!!  ;D
(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)
Title: Re: Limits
Post by Benoit_Mandelbrot on May 18th, 2004, 6:03am
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.
Title: Re: Limits
Post by Icarus on May 18th, 2004, 10:05pm

on 05/18/04 at 06:03:29, Benoit_Mandelbrot wrote:
In the case of the problem, 1^[infty] would be one.


Perhaps, but you have to prove it.
Title: Re: Limits
Post by THUDandBLUNDER on May 19th, 2004, 4:37am

on 05/17/04 at 18:50:52, TenaliRaman wrote:
QED!!  ;D
(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)

Oh, you use Graphmatica, too!

Try LIM Sin(1/x)
                        x->0
Title: Re: Limits
Post by Barukh on May 19th, 2004, 5:39am
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-.  ???
Title: Re: Limits
Post by Benoit_Mandelbrot on May 19th, 2004, 6:04am
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?
Title: Re: Limits
Post by towr on May 19th, 2004, 6:05am

on 05/19/04 at 05:39:32, Barukh wrote:
Therefore y [to] 1-.  ???
We're looking for the limit for y/(1-y)
so that would go to [infty] then..
Title: Re: Limits
Post by Barukh on May 19th, 2004, 7:42am

on 05/19/04 at 06:04:10, 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?

Yes, it's nice, but it’s not mine – you should ask TenaliRaman  ;D


on 05/19/04 at 06:05:53, towr wrote:
We're looking for the limit for y/(1-y)
so that would go to [infty] then..

So, what's the flaw in my derivation? Or, are you saying that's not enough to establish the value of y/(1-y)?
Title: Re: Limits
Post by towr on May 19th, 2004, 8:53am

on 05/19/04 at 07:42:52, 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)?
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
Title: Re: Limits
Post by TenaliRaman on May 21st, 2004, 10:45am

on 05/19/04 at 04:37:56, THUDandBLUNDER wrote:
Oh, you use Graphmatica, too!

Try LIM Sin(1/x)
                        x->0

finite value <= |1|
;)
Title: Re: Limits
Post by THUDandBLUNDER on May 23rd, 2004, 1:10pm
What is LIM[smiley=surd.gif][x(x + 1)] - x?
        x -> oo
Title: Re: Limits
Post by Barukh on May 24th, 2004, 12:06am

on 05/23/04 at 13:10:11, THUDandBLUNDER wrote:
What is LIM[smiley=surd.gif][x(x + 1)] - x?
        x -> oo

[smiley=blacksquare.gif][hide]
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].
[/hide][smiley=blacksquare.gif]
Title: Re: Limits
Post by TenaliRaman on May 24th, 2004, 11:32am
wowee,
that's elegant barukh.

much pretty straightforward method that i usually use,
::[hide]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[/hide]::
Title: Re: Limits
Post by Barukh on May 24th, 2004, 11:08pm

on 05/24/04 at 11:32:14, TenaliRaman wrote:
wowee,
that's elegant barukh.

Thanks.


Quote:
much pretty straightforward method that i usually use,

I'm having difficulty to understand your method. The first operation seems OK, but after the second one - [hide]dividing numerator by x[/hide] - you are dealing with another limit.

Or did you mean: [hide]dividing both numerator and denominator by x[/hide]?
Title: Re: Limits
Post by Sir Col on May 25th, 2004, 2:32am
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*
Title: Re: Limits
Post by Barukh on May 25th, 2004, 4:25am

on 05/25/04 at 02:32:57, 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.


Really nice!  :D
Title: Re: Limits
Post by THUDandBLUNDER on May 31st, 2004, 7:05pm
If f(x,y)= x4logexy find LIM f(x,y) as (x,y)->(0,0)
Title: Re: Limits
Post by Eigenray on May 31st, 2004, 9:32pm

on 05/31/04 at 19:05:47, THUDandBLUNDER wrote:
If f(x,y)= x4logexy find LIM f(x,y) as (x,y)->(0,0)

Try [hide]letting (x,y)->(0,0) along the curve y=e^(-1/x^n) for n=3,4,5[/hide].


Powered by YaBB 1 Gold - SP 1.4!
Forum software copyright © 2000-2004 Yet another Bulletin Board