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Can you help me evaluate this limit?

  1. Mar 9, 2008 #1
    [SOLVED] Can you help me evaluate this limit?

    lim k -[tex]\sqrt{k^{2}+k}[/tex] as k[tex]\rightarrow[/tex][tex]\infty[/tex]

    I have evaluated the limit using computer technology and I know it should be
    -1/2. I have tried using something like the squeeze theorem but failed.

    My other attempt is to mutiply the limit by the expression
    lim k +[tex]\sqrt{k^{2}+k}[/tex] / lim k +[tex]\sqrt{k^{2}+k}[/tex]

    Please help me. I know this is a simple problem, but I got stuck. Thank you
    for your time and I hope you reply.

    HF08
     
    Last edited: Mar 9, 2008
  2. jcsd
  3. Mar 9, 2008 #2
    [tex]\lim_{k\rightarrow \infty}(k-\sqrt{k^2+k})[/tex]

    [tex]\lim_{k\rightarrow \infty}\left(k-\sqrt{k^2+k} \cdot \frac{k+\sqrt{k^2+k}}{k+\sqrt{k^2+k}}\right)[/tex]

    [tex]\lim_{k\rightarrow \infty}\left(\frac{-k}{k+\sqrt{k^2+k}}\right)[/tex]
     
    Last edited: Mar 9, 2008
  4. Mar 9, 2008 #3
    Hf08

    [tex]\lim_{k\rightarrow \infty}(k-\sqrt{k^2+k})[/tex]

    This should be right...
     
  5. Mar 9, 2008 #4
    Almost there...

    Right. Now, this gives us [tex]\lim_{k\rightarrow \infty}\left(\frac{-1}{1+\frac{\sqrt{k^2+k}}{k}}\right)[/tex]

    So, the next question is how to show that [tex]\lim_{k\rightarrow \infty}\left(\frac{\sqrt{k^2+k}}{k}}\right)[/tex] = 1 . I tried l'hopital, but I wonder if there is another way?
    I think is is fairly obvious this is basically k/k for large k, but how to prove that with rigor?

    Thanks,
    HF08
     
  6. Mar 9, 2008 #5
    [tex]\lim_{k\rightarrow \infty}\left(\frac{-1}{1+\sqrt{\frac{k^2+k}{k^2}}}\right)[/tex]

    [tex]\lim_{k\rightarrow \infty}\left(\frac{-1}{1+\sqrt{1+\frac 1 k}}\right)[/tex]
     
  7. Mar 9, 2008 #6
    Thanks

    (Slaps forehead). Thanks Rocophyics. Can I modify this thread to show solved? I am still new to this forum.
     
  8. Mar 9, 2008 #7
    It's ok! Yeah, go to Thread Tools (least I think so, I never do it myself :p)
     
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