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Simple Limit Question

  1. Oct 16, 2011 #1
    Last edited: Oct 16, 2011
  2. jcsd
  3. Oct 16, 2011 #2
    Click show steps on your link. They simply rationalize the function, perform some algebra, and then use L'hopital's rule.
     
  4. Oct 16, 2011 #3
    All right, notice that you can multiply by the conjugate. So that's what we do. In essence we're multiplying by one. We multiply the top and bottom by:

    square root ((x^2)+2x-1)-x / (x^2)+2x-1)-x

    notice it's minus x, that's the conjugate. then when you simplify it you get

    (2x-1)/(square root ((x^2)+2x-1)-x

    Then you ask what's the highest power? That is x, if you said x^2, that would be wrong because notice it's the square root of x^2, so the highest power is x.

    Now you divide everything by the highest power (x), but because you're approaching negative infinity, you have to tack on a (-) negative sign in front of the bottom square root.

    now we have on top

    (2x/x) - (1/x) now evaluate as x approaches infinity...2x/x the x's cancel and we're left with 2, and 1/x...would be zero.

    so on TOP we have 2

    now to the bottom, to put something in a square root you have to square it. so on bottom we multiply by (1/x) which gets thrown into square root as dividing everything by x^2, but the -x at the end is not in the square root so we put that over x.

    on bottom we have now is -squareroot (x^2/x^2 + 2x/x - 1/x) - x/x
    evaluate as x approaches infinity so the x^2 cancel out (leaving 1), and the 2x/x^2 (becomes 0) and the 1/x^2 also becomes zero. and at the end the x/x becomes one

    what we're left with now is (-)squareroot of 1 -1, this becomes -1-1 which is -2

    and what did we have on top? 2....so now it's 2/-2 which equals -1

    that's ur limit..sorry for not using the math functions..i keep messing up on them
     
  5. Oct 16, 2011 #4

    eumyang

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    I'm assuming that you clicked the "Show steps" button and not understanding the steps shown?

    EDIT: Beaten to it. :redface:
     
  6. Oct 16, 2011 #5
    this is still the limit section, i doubt he knows what l'hopitals rule is. neither do i
     
  7. Oct 16, 2011 #6
    Don't know what Lhopital's rule is.

    I followed theclock54's post and understood everything till the part where he says "but because you're approaching negative infinity, you have to tack on a (-) negative sign in front of the bottom square root. ". I see how we get the limit when we do that, but I don't understand why we put a negative sign at the bottom.

    Thank you for the help everyone.
     
  8. Oct 16, 2011 #7

    SammyS

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    It's so handy when the expression can be read right with the post.

    lim(x→-∞) (sqrt(x2+2 x-1)+x) = -1



    [itex]\lim_{x\to\,-\infty} (\sqrt{x^2+2 x-1}+x) = -1[/itex]

    Multiply by [itex]\displaystyle \frac{\sqrt{x^2+2 x-1}-x}{\sqrt{x^2+2 x-1}-x}[/itex]

    That should give: [itex]\displaystyle \frac{x^2+2 x-1-x^2}{\sqrt{x^2+2 x-1}-x}[/itex]

    Now, multiply by (1/x)/(1/x)
    .
     
    Last edited: Oct 16, 2011
  9. Oct 16, 2011 #8
    If we were to approach positive infinity, that would mean the x's are all approaching a positive number. But the thing is when I put the x in the square root, I square it right? And if I'm approaching negative infinity, the negative just disappears because it's squared. So we must put a negative sign, to show that the x is negative. It's hard for me to explain, but after doing a few examples, you'll get what's going on.
     
  10. Oct 17, 2011 #9
    I don't really get it but next time I get a similar question I'll just remember to put in the negative sign.
     
  11. Oct 17, 2011 #10

    ehild

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    A shorter way:


    [tex]\sqrt{x^2+2x-1}=\sqrt{(x+1)^2-2}=|x+1|\sqrt{1-\frac{2}{(x+1)^2}}[/tex]

    [tex] \lim_{x \to {-\infty}}|x+1|\sqrt{1-\frac{2}{(x+1)^2}}+x= \lim_{x \to {-\infty}}
    (|x+1|+x)[/tex]

    The last step is for you... :smile:

    ehild
     
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