Mar 6, 2014 Thread starter #1 J Jamie New member Feb 11, 2014 17 I got that the limit equals 0 by simplifying the denominator from: ((x^{2}+y^{2}+1)^{1/2}) - 1 to ((x^{2} - (y+1)(y-1))^{1/2}) - 1 then ((x^{2} - (y(1+1)(1-1))^{1/2}) - 1 and then evaluating the limit by plugging in 0, getting 0/-1=0 is this correct? is there a better way to do it?

I got that the limit equals 0 by simplifying the denominator from: ((x^{2}+y^{2}+1)^{1/2}) - 1 to ((x^{2} - (y+1)(y-1))^{1/2}) - 1 then ((x^{2} - (y(1+1)(1-1))^{1/2}) - 1 and then evaluating the limit by plugging in 0, getting 0/-1=0 is this correct? is there a better way to do it?

Mar 6, 2014 Admin #2 M MarkFL Administrator Staff member Feb 24, 2012 13,775 I get a different result by converting to polar. This result is confirmed by W|A. So, I suggest using polar coordinates...what do you find?

I get a different result by converting to polar. This result is confirmed by W|A. So, I suggest using polar coordinates...what do you find?

Mar 6, 2014 Thread starter #3 J Jamie New member Feb 11, 2014 17 Could you explain in more detail? I don't think I understand what you mean

Mar 6, 2014 Admin #4 M MarkFL Administrator Staff member Feb 24, 2012 13,775 Essentially, I used $x^2+y^2=r^2$ and the limit becomes: \(\displaystyle \lim_{r\to0}\left(\frac{r^2}{\sqrt{r^2+1}-1} \right)\) And then I rationalized the denominator. You could also at this point use L'Hôpital's Rule.

Essentially, I used $x^2+y^2=r^2$ and the limit becomes: \(\displaystyle \lim_{r\to0}\left(\frac{r^2}{\sqrt{r^2+1}-1} \right)\) And then I rationalized the denominator. You could also at this point use L'Hôpital's Rule.