Problem using polar coordinates to find a limit

Question

Let $ f : \mathbb R^2 \rightarrow \mathbb R $. From the limit definition, using polar coordinates, we have that: $$ \lim \limits_{(x,y) \to (0,0)} f(x,y)=L $$ iff $ \; \forall \; ε > 0 \; \; \exists \; δ > 0 \; : \; \forall \; r,θ $ , where $ 0 < r < δ $, it is true that $$ | f(r \cos θ, r \sin θ)-L | < ε $$

Question I:

Is the following the correct negation of the above definition?

$$ \lim \limits_{(x,y) \to (0,0)} f(x,y) \neq L $$ iff $ \; \exists \; ε > 0 \; : \; \forall \; δ > 0 \; \; \exists \; r,θ $ , where $ 0 < r < δ $ , so that $$ | f(r \cos θ, r \sin θ)-L | \geq ε $$

Question II:

If it is correct, I would like to use it to prove that $ \lim \limits_{(x,y) \to (0,0)} \cfrac{x^2y}{x^4+y^2} \neq 0 $.

So, I have to find some constants $ε>0, \; θ$, such that $ A=\cfrac{r \cos^2 θ |\sin θ|}{r^2 \cos^4 θ + \sin^2 θ} \geq ε \; , \; \forall \; r>0. $

However, for $ r \rightarrow 0 $, A $ \rightarrow 0 $. So there is no such $ε$.

What am I doing wrong? Does the negation of the defintion contain any mistakes?

Thank you in advance

Answer 1

Question I: Your negation is correct, though you misplaced the phrase "such that" which should come after "$\delta >0$".

. Question II: What you want to find is some $\epsilon > 0$ such that for all $\delta>0$ you have a PAIR $r, \theta$ with $0< r= ||(r\cos \theta, r\sin \theta)||<\delta$ which satisfies $A\geq \epsilon $. I'm not entirely sure what the misunderstanding is on your end, but what you have written down in this case does not match the negation you wrote in part I.

Further, I'm not sure if you're asked to directly use the negation of the definition as part of a problem (in which case you're fine), but there are easier ways to show the limit is nonzero or does not exist. Try taking the limit along some line or curve for which you will get a nonzero value. If you have any questions on that let me know in the comments.

PS: don't be afraid to use less symbolic notation! Math is not about using notation to obscure, but to make things easier. You can often make errors by using the $\forall, \exists$ symbols too frequently. I try to avoid them if they don't make things clearer.

Answer 2

Let $f(x,y) = \frac{x^2y}{x^4 + y^2}$, then $f(r\cos(\theta),r\sin(\theta)) = \frac{r\cos^2(\theta)\sin(\theta)}{r^2\cos^4(\theta) + \sin^2(\theta)}$.

Note that $\cos^4(\theta) \in [0,1] $, so $r^2\cos^4(\theta) + \sin^2(\theta) \le r^2 +\sin^2(\theta) $, so that we have:

$|f(r\cos(\theta),r\sin(\theta)) | > \frac{r|\cos^2(\theta)\sin(\theta)|}{r^2+\sin^2(\theta)} $

Now, let's for example take $\epsilon = \frac{1}{4} $, and let $r = \theta$, then we have:

$f(r\cos(r),r\sin(r)) > \frac{r\sin(r)\cos^2(r)}{r^2+\sin^2(r)}> \frac{1}{2} \frac{\sin(r)\cos^2(r)}{r} \to \frac{1}{2} >\frac{1}{4} $ as $r \to 0^{+}$

So taking any $\delta > 0$ we can find such $ 0 <r < \delta$ ( so the pair $(r,\theta) = (r,r)$ ), such that $|f(r\cos(\theta),r\sin(\theta))| > \frac{1}{4}$

Answer 3

I think your negation is essentially correct, though I wouldn't put it that way. I think

We say that the limit of $f(x,y)$ as $(x,y)\to (0,0)$ fails to exist provided that we can find a positive $\epsilon$ such that for all $\delta>0$ with $0<r<\delta$ we fail to have $$|f(r\cos\phi,r\sin\phi)-L|<\epsilon$$ for any real $L.$

is more transparent.

As for what your main question, you formally take limits to conclude that the limit is in fact $0,$ whereas you also proved by the definition that it is not $0.$ You wish to know what's happening. Well, the problem is with the limiting operation you performed. The idea is that anytime you apply polar coordinates to take such limits, you must be sure that as $r\to 0,$ the function $f(r\cos\phi,r\sin\phi)$ converges for all $\phi.$ But here note that $$\frac{r\cos^2\phi|\sin\phi|}{r^2\cos^4\phi+\sin^2\phi} \ge \frac{r\cos^4\phi \sin^2\phi}{r^2\cos^4\phi+\sin^2\phi} \ge \frac{r\cos^4\phi \sin^2\phi}{2(r^2\cos^4\phi+\sin^2\phi)}=\frac{r\cos^4\phi \sin^2\phi}{(r\cos^2\phi+\sin\phi)^2+(r\cos^2\phi-\sin\phi)^2} \ge \frac{r\cos^4\phi \sin^2\phi}{(r\cos^2\phi+\sin\phi)^2+2(r\cos^2\phi+\sin\phi)(r\cos^2\phi-\sin\phi)+(r\cos^2\phi-\sin\phi)^2}=\frac{r\cos^4\phi \sin^2\phi}{(2r\cos^2\phi)^2}=\frac{r\cos^4\phi \sin^2\phi}{4r^2\cos^4\phi}=\frac{\sin^2\phi}{4r}.$$ Now you may note what happens as $r\to 0.$

Since it's sometimes subtle to tell immediately in polar coordinates whether a function does not converge, as in the case above, it's more convenient to use rectangular coordinates. One just has to approach the point in question differently to see that this is in fact the case. In your example, taking $y=0$ and $x\ne 0$ gives $f(x,y)=0.$ Then obviously along $y=0$ the limit at $(0,0)$ is $0.$ However, approaching along the parabola $y=x^2$ gives $f(x,y)=1/2,$ so the limiting value as we go to the origin along this parabola is $1/2.$ This clearly and more easily demonstrates the nonexistence of a limiting value at $(0,0).$