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Say, we have $f:\Bbb R^2 \to \Bbb R$ given by $$f(x,y)=\frac{x^2y}{x^2+y^2}$$

Let's try to find the double limit as $(x,y)\to(0,0)$:

Considering a path $y=mx$ the RHS evaluates to $$\frac{mx^3}{(1+m^2)x^2}$$

We need the limit as $x\to 0$, which clearly comes out to be $0$.

Now, my question is:

Does the existence of the double limit imply the function is continuous at $(0,0)$? If no, please explain why (with a counterexample if possible).

Edit:

Some people commented that the function does not exist at $(0,0)$, so it wouldn't be continuous anyway. I agree. Let us take the definition of $f$ to be as follows in that case:

$ f(x,y) = \begin{cases} \frac{x^2y}{x^2+y^2} & (x,y)\neq (0,0) \\ 0 & (x,y)=(0,0) \end{cases} $

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    $\begingroup$ Please be very specific about what you mean by double limit as $(x,y)\to (0,0)$. $\endgroup$ – Ted Shifrin Sep 13 '17 at 6:10
  • $\begingroup$ @TedShifrin Double limit is also called as simultaneous limit. We are taking $(x,y)\to(0,0)$ rather than $x\to 0 \space y \to 0$ or $y \to 0 \space x \to 0$ (which are called iterated limits or repeated limits). $\endgroup$ – user400242 Sep 13 '17 at 6:13
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    $\begingroup$ OK, so you specifically do not mean iterated limits. Then, yes, this is the usual definition of a multivariable limit. Provided $f(0,0)=\lim\limits_{(x,y)\to (0,0)} f(x,y)$, this is of course the definition of continuity at $(0,0)$. More fancily, given any $\varepsilon>0$, you're saying there is $\delta>0$ so that $\sqrt{x^2+y^2}<\delta \implies |f(x,y)-f(0,0)|<\varepsilon$. $\endgroup$ – Ted Shifrin Sep 13 '17 at 6:17
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@Blue to be clear, the existence of the limit you defined (along a path) does not mean the double limit exists. You have only found a class of relative limits. If you proved the double limit existed then all such relative limits would be equal.

Really you would need to substitute $x=r\cos \theta$, $y=r\sin\theta$: $$f(x,y) = \frac{x^2y}{x^2+y^2} = \frac{r^2\cos^2\theta\cdot r\sin\theta}{r^2\cos^2\theta + r^2\sin^2\theta} = r\cos^2\theta\sin\theta = f(r,\theta)$$ Clearly

$$0\leq |f(r,\theta)| = |r\cos^2\theta\sin\theta| \leq r$$ This is the property that allows us to control all paths that approach $(0,0)$. Suppose we have some path $\gamma(t) \to (0,0)$ as $t\to\infty$, then by controlling the radius:

$$0 \leq |f(\gamma(t),\theta)|\leq \|\gamma(t)\|_2$$ (this is just the Euclidean distance, the radius from the point at $\gamma(t)$ to $(0,0)$) we may apply the sandwich theorem as $\|\gamma(t)\|_2\to 0$, giving that $|f(\gamma(t),\theta)|\to 0$. As this held for an arbitrary path, it must hold for all of them. The limit exists.

What we have actually done is bound every single path inside shrinking disks. I get the sense that this intuition isn't really explained in typical multivariate courses.

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  • $\begingroup$ $y=mx$ cover all possible directions of linear approach towards the origin. After all $m\in \Bbb R$.In your version, $\theta$ controls the direction of linear approach as $r\to 0$. Same thing. $\endgroup$ – user400242 Sep 13 '17 at 7:11
  • $\begingroup$ @Blue They are not the same thing at all! The existence of all linear approaches is not equivalent to the existence of the double limit. What I am doing is bounding all paths inside disks with a given radius. This covers all linear paths, but it also cover quadratic paths, Bezier curves, spirals, oscillatory approaches, non-differentiable paths and even non-continuous paths. As long as the paths eventually go towards $(0,0)$, then most of the path will eventually fall within any disk, no matter how small. Being able to bound all paths in this way is the double limit. $\endgroup$ – adfriedman Sep 13 '17 at 7:29
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    $\begingroup$ No, they're different. Try this one: $$f(x, y) = \begin{cases} \frac{x^3y}{x^6+y^2},& (x, y)\neq (0, 0) \\ 0,& (x, y) = (0, 0) \end{cases}$$ All linear paths will give a directional limit of $0$ as $(x, y)\to (0, 0)$, but the function is not continuous at $(0, 0)$. Consider the path $(x, y) = (t, t^3)$ as $t\to 0$. $\endgroup$ – Michael Lee Sep 13 '17 at 7:29
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That depends on the exact definition of the double limit. Even if you take the limit simultaneously you have to consider if the value in the limit target is taken into account (the normal is not to take that value into account).

For example consider the characteristic function of the origin (being $0$ everywhere except at the origin where it's $1$). With some definitions of limits the function will have a limit of $0$ at the origin, but according to other definitions the value at origin is taken into account in which case the limit fails to exist.

If the value at the limit target is part of the definition continuity is the same as the limit exists. If that value is not part of the definition then continuity is the same as the limit exists and also is the value of the function at that point.

In your example the function is not defined at the origin which is another complication. You normally don't say that a function is continuous at a point where it's not defined. However if the limit exists we can extend the function continuously to that point.

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  • $\begingroup$ The value of the function at the origin affects continuity but never existence of the limit (unless you're in a strange foreign world where existence of a limit is defined to be continuity). Granted, the OP was discussing continuity without mentioning the value of the function at the origin. $\endgroup$ – Ted Shifrin Sep 13 '17 at 6:25
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    $\begingroup$ @TedShifrin Of course the existence of the limit is crucial if the function is defined in an neigborhood. If the limit does not exist the function is not continuous - so yes the existence affects continuity at least in some way, but with the normal definition the function need to have the same value as it's limit there. $\endgroup$ – skyking Sep 13 '17 at 6:32
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It is easy to see that $|f(x,y)| \le |y|$. Can you procced ?

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  • $\begingroup$ This should really be a comment and not an answer. $\endgroup$ – user400242 Sep 13 '17 at 7:21
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    $\begingroup$ It is an answer. My answer shows, and that should be your task: $f(x,y) \to 0$ if $(x,y) \to (0,0)$. It is now an answer in your sense ? $\endgroup$ – Fred Sep 13 '17 at 7:27

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