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.

  • 3
    $\begingroup$ 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
  • 2
    $\begingroup$ 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
  • 2
    $\begingroup$ @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

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 $y\to 0$.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.