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Simple limit at infinity

  1. Jan 23, 2007 #1
    Find the limit as x-> -infinity for (x+(x^2+12x)^1/2)

    so first of all..i multiply and divide by the conjugent then i get...

    -12x/(x-(x^2+12x)^1/2)

    i divide by x in both the nummerator and denominator to get ....
    -12/1-(1+12/x)^1/2

    so the 12/x goes to 0 and the squroot of 1 is 1 so it appears to be

    -12/1-1 which is undefined....that is not right

    where am i going wrong
     
  2. jcsd
  3. Jan 23, 2007 #2

    arildno

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    Of course it is right. The function diverges.
     
  4. Jan 23, 2007 #3
    so the limit does not exist?
    or it is undefined

    how can i express that
     
  5. Jan 23, 2007 #4

    arildno

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    It "equals" positive infinity, if you like.
     
  6. Jan 23, 2007 #5

    mathman

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    Simplify:
    (x^2+12x)^1/2 ~ x for large x, therefore your expression is ~ 2x
     
  7. Jan 23, 2007 #6

    Gib Z

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    For large x, a polynomials value is determined by its leading term. Since x is large, x^2 + 12x is x^2. That to ^1/2 is just x. x+x, 2x. Substitution, it doesn't exist.
     
  8. Jan 23, 2007 #7

    StatusX

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    The limit is as x goes to negative infinity. (x^2+12x)^1/2 looks like -x+const for x large and negative, so the series goes as x-x+const, and the limit is finite.
     
  9. Jan 23, 2007 #8

    Gib Z

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    Sorry I didnt see the negative sign....Ill rethink that
     
  10. Jan 24, 2007 #9

    dextercioby

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    The limit is -6.

    Daniel.
     
  11. Jan 24, 2007 #10

    HallsofIvy

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    Dextercioby is, as usual, correct.

    Replace the limit at [itex]-\infty[/itex] with a limit at [itex]\infty[/itex] by replacing x with -x:
    [tex]lim_{x\rightarrow -\infty} x+ (x^2+ 12x)^{\frac{1}{2}}= lim{x\rightarrow\infty} (x^2- 12x)^{\frac{1}{2}}- x[/tex]
    Multiply "numerator and denominator" by the conjugate:
    [tex]lim_{x\rightarrow\infty}\frac{x^2-12x-x^2}{(x^2-12x)^{\frac{1}{2}}+ x}= lim_{x\rightarrow\infty}\frac{-12x}{(x^2-12x)^{\frac{1}{2}}+ x}[/tex]
    Divide both numerator and denominator by x:
    [tex]lim_{x\rightarrow\infty}\frac{-12}{(1-\frac{12}{x})^{\frac{1}{2}}+ 1}[/tex]

    Now it is obvious that the numerator is -12 and the denominator goes to 2.
     
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