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