Defining a function at some given point to make it continuous

Question

So I have a function

$$ f(x,y) = \frac{xy(x^2-y^2)}{x^2+y^2} $$

I have to define the function at $(0,0)$ so that it is continuous at origin. Is there any general approach we follow in this type of problems? I am not sure but I think Sandwich Theorem has to be applied and some suitable function has to be taken .

I am not sure about the answer , please provide with some hint. I am not sure whether I selected the correct function or not — The Learner, Nov 26 '19 at 14:20
It is obvious that IF the function can be made continuous at the origin, then we need $f(0,0)=0$. For example, if we approach the origin along the $x$-axis we get 0. The harder part is proving that the limit as $(x,y)\to0$ is 0. Your answer does that. Well done! — almagest, Nov 26 '19 at 14:25

Damien · Accepted Answer · 2019-11-26T14:52:42.377

2

A possibility is to represent the point $(x,y)$ by a complex number:

$$z = x + j y = \rho e^{j \theta}$$

then

$$ xy = \rho^2 \frac{\sin 2\theta}{2} $$

$$ x^2 - y^2 = \rho^2 (2\cos^2 \theta - 1) $$

$$ x^2 + y^2 = \rho^2$$

And finally:

$$f(x, y) = \frac{xy\,(x^2-y^2)}{x^2+y^2} = f(z) = \rho^2 \frac{\sin2\theta \, (2\cos^2 \theta - 1)}{2} $$

When the point $(x, y)$ tends to $0$, $\rho$ necessarily tends to $0$ and therefore, whatever the value of $\theta$, $f(z) = f(x, y)$ tends to $0$

This method may look overskill for this simple function. The sandwich method proposed by OP works. However, OP was asking for a possible general method. This one often gives good results even in more difficult use cases.

edited Nov 26 '19 at 14:52

answered Nov 26 '19 at 14:36

Damien

1,762

Very nice method, although this question was given under Sandwich theorem category of my book . – The Learner Nov 26 '19 at 15:22
You can use the final result to easily get the expressions to use with the Sandwich method, by using bounds on $\cos(.)$ and $\sin(.)$, and replacing $\rho ^2$ by $x^2 + y^2$ – Damien Nov 26 '19 at 15:25
Yeah that's a point , thanks for the answer – The Learner Nov 26 '19 at 18:50
You welcomen happy to have helped – Damien Nov 26 '19 at 18:52

The Learner · Answer 2 · 2019-11-26T18:46:37.810

Please check out the new answer

$$ |xy(x^2-y^2)| <= |xy||(x^2-y^2)| <= |x||y||x^2+y^2|$$

Now,

$$ |x||y||(x^2+y^2)| = \sqrt(x^2)\sqrt(y^2)|x^2+y^2| <= (\sqrt{(x^2+y^2)})(\sqrt{(x^2+y^2)})|x^2+y^2| $$

The last expression is nothing but $(x^2+y^2)^2$ . Therefore

$$ \frac{ |xy(x^2-y^2)| }{(x^2+y^2)}<=\frac{(x^2+y^2)^2}{(x^2+y^2)} $$

Thus the Sandwich theorem can be applied now

Then, $$-(x^2+y^2) < f(x,y) < (x^2+y^2) $$

And I enforce the limit $(0,0)$ which gives $f(x,y)$ as $0$.

score 0 · Answer 3 · answered Nov 26 '19 at 14:16

0

I don't know whether I am right but if I take the function $(x^2+y^2)$,

Then, $$-(x^2+y^2) < f(x,y) < (x^2+y^2) $$

And I enforce the limit $(0,0)$ which gives $f(x,y)$ as $0$.

answered Nov 26 '19 at 14:16

The Learner

170

The argument is perhaps not quite convincing unless you also explain why that inequality holds. – Hans Lundmark Nov 26 '19 at 14:35
Please check out the answer above – The Learner Nov 26 '19 at 15:14
1

Looks good! (Althought it's usually preferred if you edit the old answer instead of adding a new one.) A few formatting tips: you can write \le to get $\le$ and \sqrt{x^2+y^2} to get $\sqrt{x^2+y^2}$. – Hans Lundmark Nov 26 '19 at 15:27
Thanks man , I will keep that in mind – The Learner Nov 26 '19 at 18:45

Defining a function at some given point to make it continuous

3 Answers3