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Limit Problem

  1. Apr 22, 2007 #1

    Gib Z

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    I just read the following fact and it somewhat defies my usual logic.

    [tex]\lim_{x\to\infty} \sqrt{x^2+x} - x =\frac{1}{2}[/tex].

    Now my usual logic tells me that as x becomes large, the leading term in a polynomial sequence is the major one and therefore the [itex]x^2+x[/itex] becomes [itex]x^2[/itex] so [itex]\sqrt{x^2+x}[/itex] becomes just x.

    The same logic seems to work for this limit here:https://www.physicsforums.com/showthread.php?t=166906

    However the limit is obviously not equal to zero, where is the error in this logic?
     
  2. jcsd
  3. Apr 22, 2007 #2
    divide top and bottom by x after you get rid of the radical on the top
     
  4. Apr 22, 2007 #3

    Gib Z

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    I know perfectly well how to evaluate the limit that way, Multiply by its conjugate, use difference of two squares and simplify, so on so forth. What I wanted to know was why the other method does not work.
     
  5. Apr 22, 2007 #4
    Im sure your logic will tell you that
    [tex]\sqrt{x^2+x} [/tex]
    is the same as
    [tex]\sqrt{x^2+x + \frac{1}{4} } [/tex]
    when x becomes large.
    And this is the same as
    [tex]x + \frac{1}{2}[/tex] .
     
  6. Apr 22, 2007 #5

    mjsd

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    in https://www.physicsforums.com/showthread.php?t=166906
    the logic you have just mentioned doesn't work either
    [tex]\lim_{x\rightarrow\infty} \sqrt{x^2+x+1} - \sqrt{x^2-3x}[/tex] gives you zero too if you blindly replaces all polynomials with just the term of highest degree. what makes the eventual "substitution" to work in that other example was not that you replaces everything using that logic of yours ... but to look at "limit" as x goes large, namely:

    [tex]\lim_{x\rightarrow\infty} \sqrt{x^2+x+1} - \sqrt{x^2-3x}
    = \lim_{x\rightarrow\infty} \frac{4x+1}{\sqrt{x^2+x+1} + \sqrt{x^2-3x}}[/tex]
    then divide top and bottom by x to get

    [tex]\lim_{x\rightarrow\infty} \frac{4+1/x}{\sqrt{\frac{x^2}{x^2}+\frac{x}{x^2}+\frac{1}{x^2}} + \sqrt{\frac{x^2}{x^2}-\frac{3x}{x^2}}}[/tex]

    then if you now take the limit, you get what you want.

    the lesson here is that [tex]\lim_{x\rightarrow \infty}\sqrt{x^2+x}\neq \sqrt{x^2}[/tex]
     
  7. Apr 22, 2007 #6
    It does work, you can solve this by inspection. get the conjugate and and the denominator the highest order variable is x^2, ignore the others. Walla, same at the other thread.:wink:
     
  8. Apr 22, 2007 #7

    Gib Z

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    Yes it does. In fact, I usually complete the square when questions come up like that.


    mjsd - But look at Hurkyl's comment in that thread...Sure he was not replacing the polynomials with the leading term right from the start, but he did on an equivalent expression! The fact the he couldn't do it at the start but can on an equivalent expression confuses me even more!

    O and completeing the square works for that limit as well.
     
    Last edited: Apr 22, 2007
  9. Apr 22, 2007 #8

    Gib Z

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    Yes exactly, so it does work once we have changed the form. How come it works like that, and what form does it need to be in for it to work?
     
  10. Apr 22, 2007 #9
    remember when you were first learning limits? They say you will get a fraction answer if you only focused on highest order and if the order are some on top and bottom? Are the orders same on top and bottom? you sould say yes. because limit approaches zero on all others if you rationalize the top and bottom. Also, you cannot directly evauate that limit, so you have to change its idenitity to legally find it limit that is in disguise.
     
    Last edited: Apr 22, 2007
  11. Apr 22, 2007 #10

    Gib Z

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    Ahh I see. So for this method to work, the degree of the numerator must be the same as the denominator. Ok thanks guys!
     
  12. Apr 22, 2007 #11

    Hurkyl

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    As a zeroth-order approximation, [itex]\sqrt{x^2 + x}[/itex] is equal to x, plus some terms less important than x.

    Therefore, [itex]\sqrt{x+2} - x[/itex] is equal to 0, plus some terms less important than x.


    Alas, this doesn't tell us the limit! So, we try a first-order approximation.

    [itex]\sqrt{x^2 + x} \approx x + 1/2[/itex], plus some terms less important than 1. (using the fact [itex](A + \epsilon)^n \approx A^n + nA^{n-1} \epsilon[/itex], plus terms less important than [itex]A^{n-1}\epsilon[/itex])

    Therefore, [itex]\sqrt{x+2} - x \approx 1/2[/itex], plus some terms less important than 1.

    Happily, terms less important than 1 don't matter here, so the answer is 1/2. (It would matter if, say, we had a fraction where both the numerator and denominator were zero + terms less important than 1)
     
    Last edited: Apr 22, 2007
  13. Apr 22, 2007 #12

    Gib Z

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    Where are you getting these nth order approximations from? What would the 2nd order approximation be?
     
  14. Apr 22, 2007 #13

    Hurkyl

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    Here, I'm using the binomial theorem:
    [tex]
    (A + \epsilon)^n = A^n + n A^{n-1} \epsilon +
    \frac{n(n-1)}{2} A^{n-2} \epsilon^2 + \cdots
    [/tex]
    including the fact that if A dominates [itex]\epsilon[/itex], then each term in this series dominates the tail.

    In general, you would use a differential approximation, or a Taylor series if you needed even more terms.
     
    Last edited: Apr 22, 2007
  15. Apr 22, 2007 #14
    hey Hurkle what class did you get that theorem from? I only have limited knowlege of only Calc, so I use the obvious method:redface:
     
    Last edited: Apr 22, 2007
  16. Apr 22, 2007 #15

    Gib Z

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    Umm ok but how did we know the the first order approximation didn't give us the limit and that its zero. We did you know it wasnt zero and you had to get a better approximation?
     
  17. Apr 22, 2007 #16

    Hurkyl

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    The error in the approximation [itex]\sqrt{x^2 + x} \approx x[/itex] is asymptotically less than x. Actually, because it's differentiable, we can do better: we know it's asymptotically on the order of 1 or smaller.

    So, we know the error in the approximation [itex]\sqrt{x^2 + x} - x \approx 0[/itex] is on the order of 1. So, all this tells us is when x gets big, [itex]\sqrt{x^2 + x} - x[/itex] remains bounded.


    The error in the approximation [itex]\sqrt{x^2 + x} \approx x + 1/2[/itex] is asymptotically less than 1. Actually, because it's differentiable, we can do better: we know it's asymptotically on the order of 1/x or smaller.


    So, we know the error in the approximation [itex]\sqrt{x^2 + x} - x \approx 1/2[/itex] is on the order of 1/x. So we know the limit is 1/2 as x gets big.



    Note that if we were instead looking at

    [tex]\lim_{x \rightarrow +\infty} \sqrt{4x^2 + x} - x[/tex]

    we only need the zeroth-order approximation: [itex]\sqrt{4x^2 + x} - x \approx 2x - x = x[/tex]. The error is still asymptotically on the order of 1, so x dominates, and we know the limit is [itex]+\infty[/itex].
     
  18. Apr 22, 2007 #17

    Hurkyl

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    Formally, this means
    [tex]\exists c: \forall x:|\sqrt{x^2 + x} - x| < c \cdot 1[/tex]
    Or, if you know big-Oh notation,
    [tex]\sqrt{x^2 + x} = x + O(1)[/tex]

    The less strict condition is simpler:
    [tex]
    \lim_{x \rightarrow +\infty} \frac{\sqrt{x^2 + x} - x}{x} = 0
    [/tex]
    or
    [tex]\sqrt{x^2 + x} = x + o(x)[/tex]


    And this means
    [tex]\exists c: \forall x:|\sqrt{x^2 + x} - (x + 1/2)| < c \cdot (1/x)[/tex]
    or
    [tex]\sqrt{x^2 + x} = x + 1/2 + O(1/x)[/tex]

    The less strict one is
    [tex]
    \lim_{x \rightarrow +\infty} \frac{\sqrt{x^2 + x} - (x + 1/2)}{1} = 0
    [/tex]
    or
    [tex]\sqrt{x^2 + x} = x + 1/2 + o(1)[/tex]
     
    Last edited: Apr 22, 2007
  19. Apr 22, 2007 #18
    You seem like you got the proofs down packed pretty good Hurkle. Are you a professor? Where did you learn to master proofs? Advanced Calc?
     
  20. Apr 22, 2007 #19

    Gib Z

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    I'm sorry If I seem slow, but what do you mean by asymptotically on the order of?
     
  21. Apr 22, 2007 #20

    Hurkyl

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    The intuition from asymptotic analysis (also given by nonstandard analysis) is that by asking for the limit, you want the value of the function "rounded" to the nearest constant. (or to the nearest standard number, in the nonstandard picture)

    The error in [itex]\sqrt{x^2 + x} \approx x[/itex] is on the order of a constant, which is not enough precision for the answer we desire.

    The error in [itex]\sqrt{x^2 + x} \approx x + 1/2[/itex] is on the order of 1/x, which is enough precision to get the desired answer.

    P.S. I wrote another post before your reply. :wink:
     
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