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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.

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2 Answers 2

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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.

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Hint:

Consider the curve $x=\sqrt y$ . Along this curve the limit would be

$$\lim_{y\to0}\frac{2y\sqrt y+4y^2+3y^2\sqrt y+y^3}{-2y\sqrt y-y^2}=-1$$

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  • $\begingroup$ But then $y-(x+y)^2=0$ and $f$ is undefined on that curve. $\endgroup$ Commented Jun 5, 2021 at 12:48
  • $\begingroup$ @SaikaiPrime I have edited my answer. $\endgroup$ Commented Jun 5, 2021 at 14:24

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