1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Homework Helper

    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


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    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


    User Avatar
    Homework Helper

    A shorter way:


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

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

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook