Limit exists with definition but not with polar coordinates

Question

I would like to know why if I try to prove with the delta epsilon definition that the limit as $(x,y)$ tends to $(0,0)$ is $0$ of this function: $$\frac{x^2 +y}{\sqrt{x^2+y^2}}$$ I get a positive result: for $\delta = \varepsilon -1$, I get that $|f(x,y)-0|$ is less than $\varepsilon$.
But if you evaluate the limit with polar coordinates, you get that the limit depends on the path, and thus it doesn't exist.

Answer 1

You cannot take $\delta=\varepsilon-1$, since you are after a $\delta>0$.

And you don't need polar coordinates to show that that limit does not exist. Just use the fact that, if $f(x,y)=\frac{x^2+y}{\sqrt{x^2+y^2}}$, then

• if $t>0$, $f(0,t)=1$, and therefore the limit, if it exists, can only be $1$;
• if $y=-x^2$, $f(x,y)=0$, and therefore the limt, if it exists, can only be $0$.

Answer 2

For the epsilon-delta definition you need to prove that for every positive $\epsilon$ there exists a positive $\delta$ such that...

So you can't take $\delta=\epsilon-1$, since if $0<\epsilon\leq 1$, this won't give you a positive $\delta$.