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Reveiwing for test on limits

  1. Oct 8, 2007 #1
    Im reveiwing for a test i have in about a week. Right now i have come to a problem i have a question on so here it is.

    1. The problem statement, all variables and given/known data

    lim x / x^2 - 1
    x app infinity






    3. The attempt at a solution

    lim x / x^2 - 1
    x app +infinity


    x / x ( x - 1/x )

    the x's cancel leaving

    1 / (x-1/x)

    giving

    1 / +infinity - 1/+infinity

    = 0+ ???
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 8, 2007 #2

    NateTG

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    You should probably use parentheses to clarify things...
    [tex]\lim_{x \rightarrow \infty} \frac{x}{x^2} -1=\lim_{x \rightarrow \infty} \frac{1}{x} - 1[/tex]
    [tex]\lim_{x \rightarrow \infty} \frac{x}{x^2-1}=\lim_{x \rightarrow \infty} \frac{1}{x-\frac{1}{x}}[/tex]

    Since [itex]x-\frac{1}{x}[/itex] and [itex]x[/itex] go to infinity, the fractions go to zero.
     
    Last edited: Oct 8, 2007
  4. Oct 8, 2007 #3

    CompuChip

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    Agreed

    How did you get this?
    Did you use
    [tex]\frac{1}{a + b} = \frac{1}{a} + \frac{1}{b}[/tex]
    by any chance? This is not true! For example,
    [tex]\frac{1}{4} = \frac{1}{2 + 2} \stackrel{\large\mathbf{!}}{\not=} \frac{1}{2} + \frac{1}{2} = 1[/tex]
    Instead, if you have an expression of type
    [tex]\lim_{x \to \infty} \frac{1}{f(x)}[/tex]
    with f(x) some function of x first try to determine the limit of f(x) as x goes to infinity. If f(x) goes to zero, the fraction goes to infinity (and vice versa) and if f(x) tends to a finite number F, then the limit of the fraction is 1/F.
     
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