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When I was trying to find a path that would prove that some limit doesn't exists, I was simply equaling the equation to a number and finding some expression. I will use some trivial limit, that can be easily be proven to exist by Definition or by Squeeze Theorem, to show the case. $$\lim_{(x,y)\to (0,0)}\frac{x^2y}{x^2+y^2}$$ Since this fraction is limited $$0\le\frac{x^2}{x^2+y^2}\le1$$ I can multiply both sides by $y$ $$0\lim_{(x,y)\to(0,0)}y\le\lim_{(x,y)\to (0,0)}\frac{x^2y}{x^2+y^2}\le\lim_{(x,y)\to(0,0)}y$$ Which only solution is $$\lim_{(x,y)\to (0,0)}\frac{x^2y}{x^2+y^2}=0$$ So we know the limit exists and it is equal to 0. $$ $$ But if I take this curve (that I found simply equaling the limit to 1) $$x = \sqrt{\frac{y^2}{y-1}}$$ I think it's okay because when $y\to0, x\to 0$.So the limit will be: $$\lim_{t\to 0}\frac{\frac{y^3}{y-1}}{\frac{y^2}{y-1}+y^2}=\lim_{y\to 0}\frac{\frac{y^3}{y-1}}{\frac{y^3}{y-1}}=1$$ So I found a path that proves the limit doesn't exist. But we know it exists, so must be something wrong. Did I missed something ? Where is the mistake ? I feel that there is something wrong in the domain of the curve, but since the domain is $y\gt 1$ or $y = 0$ I can't prove that with some formality.

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    $\begingroup$ You should have multiplied both sides by $|y|$ instead, but this is not a real issue because again you'll get a $0$ limit. The problem is that the path you mentioned is not defined for $y < 1$. $\endgroup$ – user258700 Jul 24 '16 at 22:17
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    $\begingroup$ You multiplied both sides by 0. This eliminated any values you wouldve had. Think x = y vs 0 = 0. 0 = 0 is the entire space whereas x = y is a subset with the possibility of being the empty set. $\endgroup$ – user64742 Jul 24 '16 at 22:19
  • $\begingroup$ @TheGreatDuck where did multiplication by $0$ both sides take place? $\endgroup$ – user258700 Jul 24 '16 at 22:22
  • $\begingroup$ Indeed you found two paths (I suspect) which have different limiting values, so the limit doesn't exist. The stuff about multiplying by $y$ and letting $y$ go to zero doesn't make sense for the reason @TheGreatDuck points out. $\endgroup$ – hardmath Jul 24 '16 at 22:25
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    $\begingroup$ @hardmath but $0 \le \frac{x^2 |y|}{x^2 + y^2} \le |y|$ does show that the limit exists and that it is equal to zero. $\endgroup$ – user258700 Jul 24 '16 at 22:37
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But if I take this curve (that I found simply equaling the limit to 1) $$x = \sqrt{\frac{y^2}{y-1}}$$

But $\varphi(y)=\left(\sqrt{\frac{y^2}{y-1}},y\right)$ is not a valid path to $(0,0)$, Namely, it is only defined for $y>1$, so you cannot follow $\varphi(y)$ while having $y\to 0$.

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  • $\begingroup$ I cant see why is not a valid path. The domain of $f(y) = \sqrt{\frac{y^2}{y-1}}$ is $y\gt 1$ or $y=0$, no ? Can you explain further ? $\endgroup$ – Jhonattan Farah Jul 24 '16 at 22:35
  • $\begingroup$ To have a valid path, you need it to be defined when $y\to 0$, i.e. when $y$ gets closer and closer to $0$ (whether it is defined for $y=0$ is irrelevant, but it matters it is defined in a neighborhood of $0$ of the form $(0,\varepsilon)$). This is not: for instance, what is $\varphi(0.1)$? $\endgroup$ – Clement C. Jul 24 '16 at 22:38
  • $\begingroup$ In other terms, you need a continuous path to $0$. $\endgroup$ – Clement C. Jul 24 '16 at 22:40
  • $\begingroup$ hm, Now I see why,Without having a continuous path to (0,0) I cant say that I am approaching the (0,0) and using it as a valid path, Which is quite obvious. Even probably there is some formality on this, I think I got it. Really thanks for the fast answer. $\endgroup$ – Jhonattan Farah Jul 24 '16 at 22:51
  • $\begingroup$ You're welcome! (If that any consolation, this is a rather sneaky mistake). $\endgroup$ – Clement C. Jul 24 '16 at 22:52

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