Defining a function at some given point to make it continuous 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 . 
 A: 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. 
A: 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 $0$.
A: 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 $0$.
