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Something's wrong

  1. Aug 14, 2009 #1
    I was given this review problem in limits.

    LimX --> inf. (1 - X + X2)1/2 - aX - b = 0.

    I was asked to find a and b such that the above equation is satisfied, which I did as follows:

    I removed an X from the entire thing and expanded the term in the brackets using binomial theorem. I get a = 1 and b = -1/2 which is the answer given in the text.

    But here's what is bothering me. I did the sum another way and got a different answer and can't put my finger on the mistake.
    I remove an X from the root term. And I split the limit across the functions since

    Lim X-->Y f(x) + g(x) = Lim X-->Y f(x) + Lim X-->Yg(x)

    Lim X-->Yf(x).g(x) = Lim X-->Yf(x) . Lim X-->Y g(x)

    This gives me

    LimX --> inf. X(1/X2 - 1/X + 1)1/2 - LimX --> inf. (aX + b) = 0

    = [LimX --> inf. X] [LimX --> inf.(1/X2 - 1/X + 1)1/2] - LimX --> inf. (aX + b) = 0

    Now the term inside the root sign goes to 1. We are left with
    LimX --> inf. X - aX - b = 0

    This way, the answer is a = 1 and b = 0.

    Why am I getting a different answer? I used all the limit rules correctly as far as I can see.
  2. jcsd
  3. Aug 14, 2009 #2
    I think you are haphazardly applying the formulas for the algebraic operations on limits. Remember, you can split limits across functions provided that all limits in question actually exist. For example, I don't think you can split the square root term as you did since X goes off to infinity.
  4. Aug 15, 2009 #3
    The second way you did it makes it so the 1 - x term disappears. Intuitively, the 1 disappearing is fine since when it comes to infinity, it's meaningless, but the x can't disappear like that.
  5. Aug 15, 2009 #4
    First, [tex] \lim_{x \rightarrow \infty} x\sqrt[]{(1/x^2 - 1/x + 1)} - ax - b = \lim_{x \rightarrow \infty} x\sqrt[]{(1/x^2 - 1/x + 1)} - (ax + b) [/tex].

    If this limit is to go to 0, then obviously a must be positive (for if a were negative the entire expression we're trying to find the limit of would increase beyond all positive bounds). But if a is positive, then the expression above is of the indeterminate form "infinity - infinity", and so you can't say that the square root expression tends to 1 and the x's will then cancel out.
  6. Aug 15, 2009 #5
    D'oh!! How could I forget that??
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