show continuity of $\frac{xy}{\sqrt{x^2+y^2}}$ How to show this function's continuity?
My book's 'Hint' says $|xy| \leq \frac12(x^2+y^2)$ can be used.
$ f(x,y) = \left\{ 
   \begin{array}{l l}
     \frac{xy}{\sqrt{x^2+y^2}} & \quad , \quad(x,y)\neq(0,0)\\
     0 & \quad , \quad(x,y)=(0,0)
   \end{array} \right.$

Here is another way to prove the continuity of $f(x,y)$ at $(0,0)$.
$$\left|\frac{xy}{\sqrt{x^2+y^2}}-0\right|={{|x||y|}\over{\sqrt{x^2+y^2}}} <{{\sqrt{x^2+y^2}}{\sqrt{x^2+y^2}}\over{\sqrt{x^2+y^2}}}=\sqrt{x^2+y^2} < \varepsilon \\\,\,(\text{where}\,\,\, \varepsilon \,\text{is a preassigned positive number})$$ if $x^2+y^2 < \delta^2,$ where $\delta =\varepsilon$. So,given any $\varepsilon >0, \exists \delta >0 $ such that $$|f(x,y)-f(0,0)|<\varepsilon \forall (x,y) \in x^2+y^2< \delta^2$$ and so $f(x,y)$ is continuous at $(0,0)$.
