Question

*calculus 3* Find the following limits if they exist

Answer 1

\[\lim_{(x,y) \rightarrow (0,0)} (x^2 + y^2)/ \sqrt{x^2+y^2+1} - 1\]

Answer 2

\[\lim_{(x,y) \rightarrow (0,0)} (x^2 + xy^2)/ (x^2 + y^2)\]

Answer 3

For the first limit, you can convert to polar using \(x^2+y^2=r^2\):
\[\lim_{(x,y)\to(0,0)}\frac{x^2+y^2}{\sqrt{x^2+y^2+1}-1}=\lim_{r\to0}\frac{r^2}{\sqrt{r^2+1}-1}\]
then apply L'Hopital's rule.

Still looking at the second limit...

Answer 4

Ah that's a good method for the first one, what I am wondering is if there is a different way to solve the limit without having to convert it into polar.

For the second limit I replaced y with x, then x^2, but I am not sure if that gives me the right answer. I got DNE

Answer 5

Replacing \(y=x\) would give
\[\lim_{(x,y)\to(0,0)}\frac{x^2+xy^2}{x^2+y^2}=\lim_{x\to0}\frac{x^2+x^3}{2x^2}=\lim_{x\to0}\frac{1+x}{2}=\frac{1}{2}\]
However, if we fix \(y=0\) we get
\[\lim_{(x,y)\to(0,0)}\frac{x^2+xy^2}{x^2+y^2}=\lim_{x\to0}\frac{x^2}{x^2}=1\]
so the limit does not exist, since \(1\neq\dfrac{1}{2}\).

Answer 6

Of course, your method works just as well. The point is that the limit is different along a different path, so you're done.

Answer 7

Thanks a bunch! :D