Can you help me evaluate this limit?

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[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
 
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[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]
 
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Hf08

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

This should be right...
 
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Almost there...

[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]
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
 
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[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]
 
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Thanks

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

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