thanks for your time and excuse me if the english is bad, it's not my first language.

Practicing for calculus 2 I found these two similar limits which have a very different result and I can't figure out the right approach to demonstrate that.

Obviously the first one (which diverges) is very easy to demonstrate if I try out some curves in the function (like for example $y=0$ and $y=x^2$). The problem is that I want to be able to solve limits with a method and not to throw random curves at them.

For example a method I've been taught to solve limits is to switch to polar coordinates. I know that polar coordinates allows me only to demonstrate the convergence of the limit but they should at least behave in a certain way that makes me see it is impossible to keep going on that path instead of make me think there may be some algebra trick I didn't try on the function.

What I am asking to you is how to recognize when I can't keep going with polar coordinates and start relying to "throwing curves at the function"

  1. $\displaystyle \lim_{(x,y)\to(0,0)}{\frac{2x^2y}{x^4+y^2}}$ diverges

In polar coordinates: $$ \lim_{\rho\to0}{\frac{2\rho^3\cos^2(\theta)\sin(\theta)}{\rho^4\cos^4(\theta)+\rho^2\sin^2(\theta)}} = \lim_{\rho\to0}{\frac{2\rho\cos^2(\theta)\sin(\theta)}{\rho^2\cos^4(\theta)+\sin^2(\theta)}} $$

I try to maximize $\displaystyle \frac{2\rho\cos^2(\theta)\sin(\theta)}{\rho^2\cos^4(\theta)+\sin^2(\theta)}$ with a function of just $\rho$ and I find two possibilities: $$ \left\lvert\, \frac{2\rho\cos^2(\theta)\sin(\theta)}{\rho^2\cos^4(\theta)+\sin^2(\theta)} \,\right\rvert \leq \left\lvert\, 2\rho\frac{1}{\sin(\theta)} \,\right\rvert $$ or $$\left\lvert\, \frac{2\rho\cos^2(\theta)\sin(\theta)}{\rho^2\cos^4(\theta)+\sin^2(\theta)} \,\right\rvert \leq \left\lvert\, \frac{2}{\rho}\sin(\theta) \,\right\rvert $$

both of which doesn't let me conclude anything.

Another thing I tried is: \begin{align} \left\lvert\, \frac{2\rho\cos^2(\theta)\sin(\theta)}{\rho^2\cos^4(\theta) + \sin^2(\theta)} \,\right\rvert &= \left\lvert\, \frac{2\rho\cos^2(\theta)\sin(\theta)}{\sqrt{\rho^2\cos^4(\theta) + \sin^2(\theta)} \sqrt{\rho^2\cos^4(\theta) + \sin^2(\theta)}} \,\right\rvert \\[4pt] &\leq \left\lvert\, \frac{2\rho\cos^2(\theta)\sin(\theta)}{\rho^2\cos^4(\theta) + \sin^2(\theta)} \,\right\rvert \\[4pt] &= \left\lvert\, \frac{2\rho\cos^2(\theta)\sin(\theta)}{\sqrt{\rho^2\cos^4(\theta)} \sqrt{\sin^2(\theta)}} \,\right\rvert \\[4pt] &= |2| \end{align} and this doesn't actually make any sense.

  1. $\displaystyle \lim_{(x,y)\to(0,0)} \frac{x^3y}{x^4+y^2}$ converges to $0$.

Again, with polar coordinates I could "maximize" $\displaystyle \frac{x^3y}{x^4+y^2}$ by the following steps $|\frac{x^3y}{x^4+y^2}|=|\frac{\rho^4 \cos^3(\theta)\sin(\theta)}{\rho^2(\rho^2\cos^4(\theta)+\sin^2(\theta))}|=|\frac{\rho^2 \cos^3(\theta)\sin(\theta)}{\rho^2\cos^4(\theta)+\sin^2(\theta)}|\leq|\frac{\rho^2 \cos^3(\theta)\sin(\theta)}{\rho^2\cos^4(\theta)}|=|\frac{\sin(\theta)}{\cos(\theta)}|$

How do I recognize that this limit requires another approach to be demonstrated while the one before actually didn't exist?

This time the square root method works and gives me $0$ as a result of the limit.


1 Answer 1


Please refer to this answer for the first part of your question. As I stated in that answer, a polar coordinate (or any coordinate) transformation is not guaranteed to demonstrate whether the limit exists. It may work, it may not--it depends on the nature of the limit in question.

  • $\begingroup$ okay but how do you explain the $|2|$ as a result and what about the second limit? $\endgroup$
    – mmAsks
    May 24, 2022 at 11:48
  • $\begingroup$ @user1189925 Although $|\sin \frac{1}{x}| \le 1$ for all $x \in \mathbb R$, does that mean $$\lim_{x \to 0} \sin \frac{1}{x}$$ is defined? Of course not. Just because a function is bounded does not mean it has a limit. The same is true here. All you have done is show that the function is bounded between $-2$ and $2$. As for your second limit, you have not provided a formal proof. You have a description of something but I don't know if it is rigorous enough to prove the existence of the limit. $\endgroup$
    – heropup
    May 24, 2022 at 14:25
  • $\begingroup$ Oh thanks, the fact that the function is bounded between 2 and -2 means nothing.But my question still stands and it is pretty concrete. When I find myself in front of an exercise like the first one and I seem to be unable to maximize the function with another function only depending from $\rho$ how do I know that I should try to demonstrate that the limit doesn't exist instead of thinking that I could try something else to demonstrate it exists like in the second exercise? I edited my post to show better what I've done in the second exercise. $\endgroup$
    – mmAsks
    May 25, 2022 at 11:51

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