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Fnd the limit of sqrt(x^2 + 4x) - x as x goes to infinity

  1. Mar 6, 2008 #1
    1. The problem statement, all variables and given/known data
    I'm a bit puzzled with a limit, as is my teacher. We came across a question that could be both factored or rationalized, but when factored, the answer is different. Perhaps someone has an explanation?

    Here it is.

    lim ((√x^2+4x)-x)
    x->∞

    I'm not going to rationalize it, because I know that is right. I'm going to show you how I'm factoring it, and perhaps someone can explain why it is not working. Thanks



    lim ((√x^2+4x)-x)
    x->∞

    lim (x(√1+4/x)-x)
    x->∞

    lim x((√1+0)-1)
    x->∞

    lim x(0)
    x->∞

    lim = 0
    x->∞

    ....
     
  2. jcsd
  3. Mar 6, 2008 #2

    dynamicsolo

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    Here's the source of the difficulty. You are starting out with what is called an "indeterminate difference", because you are subtracting infinity from infinity, which can give you anything at all. By factoring out the "x", the other factor is now

    (?1+4/x) - 1 ,

    which, for the limit at infinity, is indeed zero. The trouble is that your full expression now gives a limit of infinity times zero, which is called an "indeterminate product", as this can also give you anything at all.

    The most direct way of dealing with such indeterminate differences is to use the "conjugate factor" method (that's what I call it -- I don't know its "official" name, or if it even has one). Multiply your expression by the ratio

    ((?x^2+4x) + x) / ((?x^2+4x) + x) , which is just one. This gives you the quotient

    [((?x^2+4x) - x) · ((?x^2+4x) + x)] / ((?x^2+4x) + x) .

    The factors in the numerator are the factors of a difference of two squares, which is to say that if you multiply those two binomials, the "cross terms" will cancel. You now have

    [(x^2+4x) - (x^2)] / ((?x^2+4x) + x) , which simplifies to

    (4x) / ((?x^2+4x) + x) .

    The point of this method is that it eliminates places where there are differences between terms, which lead to problems in taking limits. We still want the limit at infinity, so divide numerator and denominator here by x , which leads to

    4 / ((?1 + 4/x) + 1) ,

    for which the limit at infinity is 4 / [(?1+0) + 1] = 2 . Putting your original function into a graphing routine will confirm that it has a horizontal asymptote of y = 2.

    EDIT: Either the board or my browser doesn't like the square root symbol you used when I put it in my reply. So, if you're seeing ?'s all over the place, those are supposed to be radicals...
     
    Last edited: Mar 6, 2008
  4. Mar 6, 2008 #3

    quasar987

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    lim (x(√1+4/x)-x)
    x->∞

    lim x((√1+0)-1)
    x->∞

    why did that 4/x become 0? you can't just plug the limit in one of the factor like that because what you've got there is an indeterminate form ∞0.
     
  5. Mar 6, 2008 #4
    I was taught to plug a zero in wherever x is in the denominator. ie. 4/x = 0 when dealing with limits of infinity
     
  6. Mar 6, 2008 #5
    is there any way to tell when you've got a function in indeterminate form before factoring?
     
  7. Mar 6, 2008 #6

    dynamicsolo

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    The first thing to do with a limit problem is just follow the "limit laws" and see what comes out. If you can follow proper rules of arithmetic to get an answer (including infinity), then that result is the limit. The following results are "indeterminate" and require some other approach to be used to evaluate the limit properly:

    0/0 , [tex]\frac{\infty}{\infty}[/tex] (indeterminate ratios)
    [tex]0 · \infty[/tex] (indeterminate product)
    [tex]\infty - \infty[/tex] (indeterminate difference)
    [tex]0^{0} , 1^{\infty}[/tex] (indeterminate powers)

    That is true by the "reciprocal powers" limit law: the limit at positive or negative infinity of a real number divided by a positive real power of x is zero. Where you had applied it is fine; the difficulty for that limit lay elsewhere. You have to look over the behavior of the terms in the function in the infinite limit to see if there will be trouble...
     
    Last edited: Mar 6, 2008
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