Can you help me evaluate this limit?

[HF08]

[SOLVED] Can you help me evaluate this limit?

lim k -$$\sqrt{k^{2}+k}$$ as k$$\rightarrow$$$$\infty$$

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 +$$\sqrt{k^{2}+k}$$ / lim k +$$\sqrt{k^{2}+k}$$

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

 

[rocomath]

$$\lim_{k\rightarrow \infty}(k-\sqrt{k^2+k})$$

$$\lim_{k\rightarrow \infty}\left(k-\sqrt{k^2+k} \cdot \frac{k+\sqrt{k^2+k}}{k+\sqrt{k^2+k}}\right)$$

$$\lim_{k\rightarrow \infty}\left(\frac{-k}{k+\sqrt{k^2+k}}\right)$$

 

[HF08]

Hf08

$$\lim_{k\rightarrow \infty}(k-\sqrt{k^2+k})$$

This should be right...

 

[HF08]

Almost there...

$$\lim_{k\rightarrow \infty}(k-\sqrt{k^2+k})$$

$$\lim_{k\rightarrow \infty}\left(k-\sqrt{k^2+k} \cdot \frac{k+\sqrt{k^2+k}}{k+\sqrt{k^2+k}}\right)$$

$$\lim_{k\rightarrow \infty}\left(\frac{-k}{k+\sqrt{k^2+k}}\right)$$

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

So, the next question is how to show that $$\lim_{k\rightarrow \infty}\left(\frac{\sqrt{k^2+k}}{k}}\right)$$ = 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

 

[rocomath]

$$\lim_{k\rightarrow \infty}\left(\frac{-1}{1+\sqrt{\frac{k^2+k}{k^2}}}\right)$$

$$\lim_{k\rightarrow \infty}\left(\frac{-1}{1+\sqrt{1+\frac 1 k}}\right)$$

 

[HF08]

Thanks

$$\lim_{k\rightarrow \infty}\left(\frac{-1}{1+\sqrt{\frac{k^2+k}{k^2}}}\right)$$

$$\lim_{k\rightarrow \infty}\left(\frac{-1}{1+\sqrt{1+\frac 1 k}}\right)$$

(Slaps forehead). Thanks Rocophyics. Can I modify this thread to show solved? I am still new to this forum.

 

[rocomath]

It's ok! Yeah, go to Thread Tools (least I think so, I never do it myself :p)