I have the following limit:

$\lim \limits_{(x, y) \to (0,0)} \cfrac{x^2y^2}{x^2+y^4}$

According to my book the limit is supposed to exist and be equal to 0, but I'm not sure what I'm doing wrong.

First, I try to approach it from two different directions, I'll skip this and say that I tried 3 different directions and got 0 everytime. Now onto polar coordinates:

$\lim \limits_{\rho \to 0} \cfrac{\rho cos^2 \theta \rho^2sin^2\theta}{\rho^2cos^2\theta+\rho^4sin^4\theta} = \lim \limits_{\rho \to 0} \cfrac{\rho^4cos^2\theta sin^2\theta}{\rho^2(cos^2\theta+\rho^2sin^4\theta)}$

Simplifying the above expression:

$\lim \limits_{\rho \to 0} \cfrac{\rho^2cos^2\theta sin^2\theta}{cos^2\theta+r^2sin^4\theta}$

Now, when $\rho \to 0$ the above function is dependant on theta, and even though the numerator tends to $0$ there's the chance of $\theta = k\pi-\cfrac{\pi}{2}$ meaning that we'd end up with $cos^2\theta = 0$ therefor we'd be in a $0/0$ situation. Because of that, the above limit is dependant on $\theta$ and not $\rho$, and that means that the function will assume different values on different directions, meaning that the limit does not exist, but according to my textbook the limit is 0. I really can't understand what I'm doing wrong.

  • $\begingroup$ Hint: you have 2 cases: the limit when $\cos \theta \neq 0$, which is easy to compute; and the case when $\cos \theta = 0$, which is also easy to compute. $\endgroup$ – Ertxiem - reinstate Monica Mar 29 '19 at 12:19
  • $\begingroup$ I am writing an answer, I request you to wait. $\endgroup$ – Teresa Lisbon Mar 29 '19 at 12:30

Up till the quotient $\frac{\rho^2 \cos^2 \theta\sin^2 \theta}{\cos^2 \theta + \rho^2\sin^4 \theta}$ you are fine.

Now when $\rho \to 0$ the above function is dependent on $\theta$.

All right, but the dependency is not good enough to prevent the limit being zero!

there is a change of a $0/0$ at $k\pi - \frac{\pi}{2}$.

Indeed, note that at such values of $\theta$, the denominator is not zero because the sine term is not zero! Whatever path we approach $(0,0)$ from, the denominator cannot be zero, because one of the two terms will always be non-zero.

The function will assume different values in different directions, meaning that the limit does not exist.

Once again, not different enough to prevent convergence to zero. Yes, it behaves "very differently" if we approach along the lines $\theta = 15^\circ$ and $75^\circ$, but it doesn't matter : convergence to zero will happen.

The point is, that you haven't done anything wrong, but just panicked, thinking that the function could do different things for different $\theta$. Yes, it does do different things, but near $(0,0)$ those things aren't far off each other. What's the reason?

What you've got to see is this :what is the dominant term of the expression? Which term of the expression exerts the most influence on the expression when your point is close to $(0,0)$?

Well, $\sin \theta$ and $\cos \theta$ are always confined to $[-1,1]$, so their behaviour is always limited, in that you can always bound their behaviour by the constants $-1$ and $1$. Let us use this fact.

How do we use it? Let us make a small transformation, where we take the $\rho^2$ to the denominator. $$ \frac{\rho^2 \cos^2 \theta \sin^2 \theta}{\cos^2 \theta + \rho^2 \sin^4 \theta} = \frac{\cos^2 \theta \sin^2 \theta}{\sin^4 \theta + \color{blue}{\rho^{-2}}\cos^{2} \theta} $$

Now, we look at the numerator. Regardless of the value of $\theta$, we have that the numerator will be between $-1$ and $1$. So, the numerator doesn't really move here : all the big stuff is in the denominator.

In the denominator, the point is that $\mathit \rho$ is super super small, so $\rho^{-2}$ is super super big. Therefore, if $\sin \theta \neq 0$, the contribution of $\rho^{-2}\sin^4 \theta$ can be made much much larger than the contribution of $\cos^2 \theta$ by simply taking $\rho$ small enough.

If $\sin \theta = 0$ then $\cos \theta = 1$ so the denominator is $1$, but then the numerator is zero, so the given expression is zero.

Either way, you see , at least geometrically, that independent of $\mathit \theta$, the given expression behaves very much like $0$ for $\mathit \rho$ chosen sufficiently small.

Of course, other answers will suggest how to approach this rigorously, but your takeaway should be this : as long as there's a "controlling/dominating term" (imagine a policeman trying to prevent people getting out of their houses during a curfew), there's not much that varying $\theta$ can get you : the misbehaviour of the changing $\theta$ is only so much.

Which, of course, is also the content of the famous squeeze principle or sandwich theorem of limits. From a rigorous point of view, the answer to these sort of questions is to identify bounds for the non-dominating terms, and use the squeeze principle, as the others have done.

  • $\begingroup$ Thank you very much for your detailed response! $\endgroup$ – Frost832 Mar 29 '19 at 12:47
  • $\begingroup$ You are welcome! Please read carefully the details : I like the fact that you have seen this limit geometrically and then deduced it did not exist , because it allows me to correct you and show you the geometric picture better. $\endgroup$ – Teresa Lisbon Mar 29 '19 at 12:49
  • $\begingroup$ I saved your comment, it was an amazing explanation. Thanks again :) $\endgroup$ – Frost832 Mar 29 '19 at 13:05

If $\cos^{2}(\theta)=0$ then $sin^{2}(\theta) =1$ so you don't have a $\frac 0 0$ situation.

Proof that the limit is $0$: $2xy^{2}\leq x^{2}+y^{4}$ so $|\frac {x^{2}y^{2}} {x^{2}+y^{4}}|\leq \frac 1 2 |x| \to 0$.


Polar coordinates aren't necessary here. You may use the inequality

  • $\sqrt{ab} \leq \frac{a+b}{2}$ for $a,b\geq 0$
  • $\Rightarrow x^2 + y^4 \geq 2\sqrt{x^2y^4}=2|x|y^2$.
  • Note that if either $y= 0$ or $x= 0$ the expression is equal to zero anyways.

So, let's assume $xy \neq 0$:

$$\frac{x^2y^2}{x^2+y^4}\leq \frac{x^2y^2}{2|x|y^2}=\frac{|x|}{2} \stackrel{(x,y)\to (0,0)}{\longrightarrow}0$$


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