A strange confusion over a problem of continuity in Multivariate Calculus.

Question

For $\beta\in\Bbb R,$ define $$f(x,y)= \begin{cases}\cfrac{x^2|x|^{\beta}y}{x^4+y^2},&x\neq 0 \\ 0, &x=0 \end{cases}$$

Prove that at $(0,0)$ the function is discontinuous if $\beta=0.$

My solution goes like this:

We can prove that $f$ is discontinuous at $(0,0)$ if, $\lim_{(x,y)\to(0,0)}f(x,y)\neq f(0,0).$

We try to show that in this case, $\lim_{(x,y)\to(0,0)}f(x,y)$ does not exist.

A general strategy to show this, is to take two functions, say $y=\phi(x)$ and $y=\psi(x)$ such that $\lim_{x\to 0}\phi(x)=\lim_{x\to 0}\psi(x)=0$ and show that $\lim_{x\to 0} f(x,\phi(x))\neq \lim_{x\to 0} f(x,\psi(x))$ (, also we note that, as $x\to 0$ we have, $(x,y)=(x,\phi(x))\to(0,0)$ and $(x,y)=(x,\phi(x))\to (0,0)$).

We let $y=\phi(x)=x$ and $y=\psi(x)=x^2.$ We have, $$f(x,y)=\begin{cases} \cfrac{x^2y}{x^4+y^2}, & x\neq 0 \\ 0,&x=0 \end{cases}$$ and so, if $L=\lim_{(x,y)\to (0,0)} f(x,y)$ exists then, $L=\lim_{x\to 0}f(x,\phi(x))=\lim_{x\to 0}\frac{x^3}{x^4+x^2}=\lim_{x\to 0}\frac{x}{x^2+1}=0$.

Again, $L=\lim_{x\to 0}f(x,\psi(x))=\lim_{x\to 0}\frac{x^4}{x^4+x^4}=\lim_{x\to 0}\frac{1}{2}=\frac 12.$

This is what we've wanted to show and $L=\frac 12$ gives us a contradiction as $\frac 12\neq 0.$

So, $\lim_{(x,y)\to(0,0)}f(x,y)$ does not exist. This proves that at $(0,0)$ the function is discontinuous if $\beta=0.$

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On second thought, I tried to prove the same thing using the "approach of polar coordinates' as follows:

Just like the previous approach, we try to show that $\lim_{(x,y)\to(0,0)}f(x,y)$ does not exist but this time using polar coordinates.

We have,

$\lim_{(x,y)\to(0,0)}\cfrac{x^2y}{x^4+y^2}=\lim_{r\to 0} \cfrac {r^3\cos^2\theta\sin\theta}{r^4\cos^4\theta+r^2\sin^2\theta}=\lim_{r\to 0} \cfrac {r\cos^2\theta\sin\theta}{r^2\cos^4\theta+\sin^2\theta}=0.$

So, surprisingly we end up seeing that $\lim_{(x,y)\to(0,0)}f(x,y)=f(0,0)$ due to which we may conclude that $f$ is continuous at $(0,0).$

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Both approaches should have given me the same conclusion, but this is not happening in my case. However, I don't understand where did I go wrong in this second approach of mine? Is my first approach, correct? Any help regarding this issue will be greatly appreciated.

Answer 1

From the purely logical standpoint:

The equality \begin{equation} \tag{$1$} \lim_{(x, y) \to (0, 0)} g(x, y) = \ell \end{equation} holds if and only if the equality \begin{equation} \tag{$2$} \lim_{t \to 0} g(\varphi(t), \psi(t)) = \ell \end{equation} holds for every continuous curve $(x, y) = (\varphi(t), \psi(t))$ with $(\varphi(0), \psi(0)) = (0, 0)$.

• In your first approach you found two curves that gave different values of the limit ($2$). This correctly proves that the limit ($1$) does not exist.

• In your second approach you essentially proved that when the curve $(\varphi(t), \psi(t))$ is any straight line, the value of the limit ($2$) equals $0$. This does not prove that the limit ($1$) equals $0$, because you haven't considered every possible curve.

The reason why your approach with polar coordinates is equivalent to computing the limit over a straight line (rather than an arbitrary curve) is as follows: the limit

$$\lim_{r \to 0} g(r \cos \theta, r \sin \theta)$$

is a special case of the limit ($2$), where $\varphi(r) = r \cos \theta$ and $\psi(r) = r \sin \theta$. The curve that it gives, i.e. $\{ (\varphi(r), \psi(r)) : r \in \mathbb{R} \}$ is a straight line going in the direction of $(\cos \theta, \sin \theta)$.

Answer 2

Converting to polar coordinates does not alter the fact that, along the curve $y=x^2$, the fraction has constant value equal to $\frac{1}{2}$.

If you liked you could substitute polar coordinates to convert the equation of the curve $y=x^2$ to $r \sin \theta = r^2 \cos^2 \theta$; when you cancel the $r$ that becomes $\sin\theta =r \cos^2 \theta$. Substituting this into the polar coordinate fraction, the numerator becomes $r^2 \cos^4 \theta$ and the denominator becomes $2r^2 \cos^4 \theta$ and so the whole thing simplifies to a constant value of $\frac{1}{2}$ along that curve, just as it did in Cartesian coordinates.

So the lesson remains the same: just because you are letting $r \to 0$ you are not free to "hold $\theta$ fixed". You have to consider the limit along all possible ways to let $r$ approach zero, even along curves like $\sin\theta = r \cos^2\theta.$

Answer 3

For the "polar test" to be conclusive in showing that $\lim_{(x, y) \to O} f (x, y) = L$, two conditions together are necessary (and sufficient):

(1) For each $\theta \in [0, 2 \pi)$, the limit $\lim_{r \to 0} f (r \cos \theta, r \sin \theta)$ must exist and equal $L$.

(2) The limit in (1) is uniform in $\theta$. That is, for any $\varepsilon > 0$, there must exist some $\delta$ such that $|f (r \cos \theta, r \sin \theta) - L| < \varepsilon$ for $r < \delta$ and all $\theta \in [0, 2 \pi)$.

This fact is proven nicely here.

In your case, (1) is certainly satisfied, since for $\theta \ne 0$,

$$\lim_{r \to 0} f (r \cos \theta, r \sin \theta) = \lim_{r \to 0} \frac{r \cos^2 \theta \sin \theta}{r^2 \cos^4 \theta + \sin^2 \theta} = 0$$

and for $\theta = 0$, the limit becomes $\lim_{r \to 0} \frac{0}{r^2} = 0$.

This is all you checked, so with just (1), the continuity of $f$ at $O$ is indeed inconclusive, as I mentioned.

Your situation fails to satisfy (2): the limit is not uniform in $\theta$. This you have effectively already shown. For any fixed $r > 0$, the value of $f (r \cos \theta, r \sin \theta)$ can always be made to be $\frac{1}{2}$ when $\cos \theta \cot \theta = \frac{1}{r}$. So, no $\delta$ exists for $\varepsilon < \frac{1}{2}$, showing that the limit $\lim_{(x, y) \to O} f (x, y)$ is not $0$.