# The limit $\lim\limits_{(x,y)\to (0,0)}(x+y)\frac{y+(x+y)^2}{y-(x+y)^2}$ does not exist.

If $$f(x,y)=(x+y)\cdot \frac{y+(x+y)^2}{y-(x+y)^2}$$ show that $$\displaystyle \lim_{(x,y)\to(0,0)} f(x,y)$$ does not exist.

Since both the iterated(repeated) limits exist and are 0, if the double limit exists it must be equal to 0. Hence I have to prove the double limit can't be 0.

I plotted $$f(x,y)=0.5$$ on Desmos and saw a portion of the curve is arbitrarily close to the origin but not continuous there. So no matter how small $$\delta>0$$ you choose, $$f(x',y')=0.5>\epsilon$$ for some $$|x'|<\delta,|y'|<\delta$$. Hence the double limit doesn't exist.

But since the question is from Analysis course, I must solve it analytically. Can anyone define $$x=\phi(y)$$ or $$y=\psi(x)$$ s.t. their limits are different for different constants used ! Although you can solve it with different approaches also.

First substitute $$t=x+y$$ to get $$\lim_{(x,y)\to(0,0)}f(x,y) = \lim_{(t,y)\to(0,0)}t\cdot\frac{y+t^2}{y-t^2}$$ Now substituting $$y=kt^3+t^2$$ gives the limit $$\frac2k$$, for any constant $$k$$. So limit doesn't exist.
Consider the curve $$x=\sqrt y$$ . Along this curve the limit would be
$$\lim_{y\to0}\frac{2y\sqrt y+4y^2+3y^2\sqrt y+y^3}{-2y\sqrt y-y^2}=-1$$
• But then $y-(x+y)^2=0$ and $f$ is undefined on that curve. Commented Jun 5, 2021 at 12:48