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A limit problem

by suffian
Tags: limit, solved
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suffian
#1
Jun2-03, 12:19 PM
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I just need some help showing how this limit systematically follows from the limit rules:

           # x     2
      1   #       t           1
lim  ---  #  ---------- dt = ---
x->0   3  #     2             3
      x  # 0   t  + 1
My first chain of thought led to breaking the expression up as follows:
1/x^2 * ( 1/x * Integral[0..x, t^2/(t^2+1)] )

Then I just kind of figured that the subexpression on the right was the average value of the function being integrated from 0..x and as x->0 the average value would approach x^2/(x^2+1), which led to:

1/x^2 * x^2/(x^2+1) = 1/(x^2+1)
which would approach one as x approached zero.

But clearly that's wrong (not surprisingly since I made a sketchy move in the middle) since the answer is one-third. Can anyone show me how to do this?

edit: possibly w/o actually integrating because this is an exercise in which you're expected to know the FTofC but not how to integrate that.

edit2: oh, not supposed to no l'hospital's rule either.
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Hurkyl
#2
Jun2-03, 04:00 PM
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Your first thought would have worked if you only had 1/x out front instead of 1/x3. (because your limit would essentially be a derivative)

So the trick is to rewrite it in a form in which the fundamental theorem of calculus applies! In particular, if you can do a substitution in the integral so the bounds of integration are from 0 to x3, then you can use your thought to evaluate the limit.
suffian
#3
Jun2-03, 05:34 PM
P: n/a
Okay, I think this works then.

I tried to change Integral[0..x, t2/(t2+1)] into Integral[0..x3, f(t)]

Integral[0..x, t2/(t2+1)] = Integral[0..x3, f(t)]
d/dx[ Integral[0..x, t2/(t2+1)] ] = d/dx[ Integral[0..x3, f(t)] ]
x2/(x2 + 1) = f(x3).3x2
f(x3) = 1/(3(x2+1)), (x != 0)
f(x) = 1/(3(x2/3+1))

so..
Limit[ x->0, 1/x3 Integral[0..x3, 1/(3(t2/3+1))] ]

Which is the average value of the function inside the integral from 0..x3, which approaches f(0) as x approaches zero, which would be 1/3. I hope that's a sufficient way to solve the problem.

Hurkyl
#4
Jun2-03, 08:42 PM
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A limit problem

That's an interesting way to change the limits! But you got the right answer in the end, I'm gonna have to look at it and see why it works; I never thought to do it that way.

Incidentally, I was thinking doing the substitution t = s1/3. t = x => s = x3, t = 0 => s = 0, and the integral became

∫0..x3 s2/3 / (s2/3 + 1) * (1/3) s-2/3 ds

which is precisely the integral you got.


Anyways, then you're left with the form:

L = limx->0 1/x3 &int0..x3 g(s) ds

We are also permitted to substitute in the limit variable, and I will do so to make things simpler. x3 = y

L = limy->0 1/y &int0..y g(s) ds

By the fundamental theorem of calculus, if G(s) is the antiderivative of g(s):

L = limy->0 (G(y) - G(0)) / y = G'(0) = g(0)

So that's how you rigorously justify your last step.
suffian
#5
Jun2-03, 11:15 PM
P: n/a
Thanks!
Hurkyl
#6
Jun2-03, 11:20 PM
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I guess, for the sake of completeness, I should specify that when I changed the limit variable, I had to use a function that is continuous and invertible near x = 0 (I think that alone is sufficient to permit the operation). y(x) = x1/3 satisfies that condition.


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