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Math Help - New Limit questions

  1. #1
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    Exclamation New Limit questions

    how do you solve the question in the picture?

    Moderator edit: The question is \lim_{x \rightarrow - \infty} (x + \sqrt{x^2 + 3x}).
    Attached Thumbnails Attached Thumbnails New Limit questions-math.jpg  
    Last edited by mr fantastic; October 6th 2009 at 03:47 PM.
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  2. #2
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    Quote Originally Posted by Sneaky View Post
    [IMG]file:///C:/DOCUME%7E1/HP_Owner/LOCALS%7E1/Temp/moz-screenshot.png[/IMG]how do you solve the question in the picture?

    First we multiply the expression by 1:

    [x + Sqrt(x^2 + 3x)]*(x - Sqrt(x^2 + 3x)]/[x - Sqrt(x^2 + 3x)] =

    [x^2 - x^2 - 3x]/[x - Sqrt(x^2 + 3x)] = -3x/[x - Sqrt(x^2 + 3x)]

    Now we're going to multiply the above by 1 = (1/-x)/(1/-x)...why -1/x?
    Because we're going to "put" 1/-x inside the square root in the denominator, but as x --> -oo we HAVE to put inside the root something positive! Said this, pay atention to the fact that 1/-x gets into the root
    as 1/x^2 , so :

    -3x/[x - Sqrt(x^2 + 3x)]*(1/-x)/(1/-x) = 3/[-1 - Sqrt(1 + 3/x)] --> 3/-2

    by arithmetic of limits and since 3/x --> 0 when x --> +/- oo.

    You can now grab your personal calculator and input low values for x into the function, say x = -1,000 or x = -100,000, and convince yourself the limit is -3/2 indeed.

    Tonio
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  3. #3
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    i dont understand

    -3x/[x - Sqrt(x^2 + 3x)]*(1/-x)/(1/-x) = 3/[-1 - Sqrt(1 + 3/x)] --> 3/-2

    that line at all

    how do u get the -3/2
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  4. #4
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    Quote Originally Posted by Sneaky View Post
    i dont understand

    -3x/[x - Sqrt(x^2 + 3x)]*(1/-x)/(1/-x) = 3/[-1 - Sqrt(1 + 3/x)] --> 3/-2

    that line at all

    how do u get the -3/2

    Hmmm...3/x --> 0 and thus 1 + 3/x --> 1 and thus Sqrt(1 + 3/x) --> Sqrt(1) = 1.

    In conclusion 3/[-1 - Sqrt(1 + 3/x)] --> 3/(-1 - Sqrt(1)) = 3 /(-1-1) = 3/-2

    Tonio
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  5. #5
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    i got to the step

    -3x (1/-x)
    -------------------------
    (x- sqrt(x^2 + 3x))(1/-x)

    then for the numerator i simplify and get 3
    but for the bottum part i dont see how you get -2
    and why does 3/x = 0, why do you assume that?

    why does

    -sqrt(x^2 + 3x)
    ---------------- = -1 ?
    -x
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  6. #6
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    ok so
    sqrt(x^2 + 3x)
    ----------------
    x
    with this i understand how to get one, so now i get

    3
    -------
    -1 +/- 1

    so its either -3/2 or 3/0

    so how do you know its -3/2 not the other one?
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  7. #7
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    Quote Originally Posted by Sneaky View Post
    i got to the step

    -3x (1/-x)
    -------------------------
    (x- sqrt(x^2 + 3x))(1/-x)

    then for the numerator i simplify and get 3
    but for the bottum part i dont see how you get -2
    and why does 3/x = 0, why do you assume that?

    Oh, dear god: who ever wrote such a nonsense?! I said 3/x --> 0, not equality! Meaning, of course, that x--> oo or x --> -oo, it's the same in both cases.

    Now the denominator:

    (x- sqrt(x^2 + 3x))(1/-x) = x*(-1/x) - (-1/x)Sqrt(x^2 + 3x) =

    -1 - Sqrt([x^2 + 3x]/x^2) = -1 - Sqrt(1 + 3/x)

    Rememebr that if a is a non-negative real number, the a*Sqrt(h) = Sqrt(ah) , and this is why we needed -1/x = 1/-x and NOT 1/x, because since
    we're taking the limit as x --> -oo we're going to work with NEGATIVE x's, so in order to get somethinbg into a square root we took -1/x, which is POSITIVE (as 1/x is negative!).

    Tonio

    Pd. Read carefully this and my prior answers, work on them with pencil and paper, THINK...and after a good while of honest effort, if you don't succeed then write back.


    why does

    -sqrt(x^2 + 3x)
    ---------------- = -1 ?
    -x
    ---------------------------------------------
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