Is $\frac{x^2y^2}{\sqrt{x^2+y^2}}$ continuous? Let $f: \mathbb{R}^2 $ -> $\mathbb{R}$
I was wondering if this function is a continuous function. 

Can I just say that $\frac{x^2y^2}{\sqrt{x^2+y^2}}$ is continuous everywhere except perhaps at $x=0$, because $\lim_{x\rightarrow \infty}$ $\frac{x^2y^2}{\sqrt{x^2+y^2}}$ = $\infty$, and the definition is that if $\lim_{x\rightarrow c}$ $f(x)$ = $f(c)$ then $f(x)$ is continuous at $c$. 
 A: We know that if $f$ and $g$ are continuous functions then $f(x,y)\cdot g(x,y)$ is continuous and if $g(x,y)\neq 0$ then $\frac{f(x,y)}{g(x,y)}$ is continuous. Using this what can you say about continuity outside $0$?
If we want to check $0$ independently we need to show that
$$\lim_{(x,y)\rightarrow 0}f(x,y) = 0.$$
The easiest way of doing this would be using polar coordinates if you are familiar with them? Write $x = r\cos \phi$ and $y = r\sin \phi$ and see what happens when $r\rightarrow 0$. The intuition of this is given by the following image:

In your case we would get that
$$f(r\cos \phi,r\sin \phi) = \frac{r^4\cos^2\phi\sin^2\phi}{\sqrt{r^2(\cos^2\phi+\sin^2\phi)}}$$
if you simplify this maybe it will become clear that this does not tend to infinity as $(x,y)\rightarrow 0\Leftrightarrow r\rightarrow 0$. I hope this helps
A: Use that $$\sqrt{x^2+y^2}\geq \sqrt{2|xy|}$$ then is $$\frac{x^2y^2}{\sqrt{x^2+y^2}}\le \frac{x^2y^2}{2^{1/2}|xy|^{1/2}}$$
A: Note that by polar coordinates
$$\frac{x^2y^2}{\sqrt{x^2+y^2}}=r^3 \cos^2\theta \sin^2\theta \to 0$$
A: The function $f$ is continuous in $\mathbb{R}^2\setminus\{(0,0)\}$ because it is the quotient of two cntinuous functions there.
On the other hand, if $(x,y)\neq(0,0)$, then$$f(x,y)=\frac{x^2y^2}{\sqrt{x^2+y^2}}\leqslant\frac{\sqrt{x^2+y^2}^2\sqrt{x^2+y^2}^2}{\sqrt{x^2+y^2}}=\sqrt{x^2+y^2}^3$$and therefore$$0\leqslant\lim_{(x,y)\to(0,0)}f(x,y)\leqslant\lim_{(x,y)\to(0,0)}\sqrt{x^2+y^2}^3=0.$$So$$\lim_{(x,y)\to(0,0)}f(x,y)=0=f(0,0),$$which means that $f$ is continuous at $(0,0)$ too.
A: Let $|x| <1$, $|y|<1$.
$|f(x,y)| =|\dfrac{x^2y^2}{\sqrt{x^2+y^2}} | \lt$
$\dfrac{|x||y|}{\sqrt{x^2+y^2}} =$
$\dfrac{\sqrt{x^2}\sqrt{y^2}}{\sqrt{x^2+y^2}}\le$
$\dfrac{\sqrt{x^2+y^2}\sqrt{x^2+y^2}}{\sqrt{x^2+y^2}}=$
$\sqrt{x^2+y^2} .$
Choose $\delta = \epsilon.$
