Corresponding continuous functions to two variable functions I am new to this topic, and I have this question that I am not really sure how to approach. Any help would be appreciated.
Decide for each of the following functions $f, g, h: \mathbb{R}²$\{$(0,0)$}$  : \rightarrow \mathbb{R}$, if there exists a continuous function $\tilde{f}$, $\tilde{g}$, $\tilde{h}$: $\mathbb{R}²  \rightarrow \mathbb{R}$ which correspond with $f, g,h$ respectively on $\mathbb{R}²$\{$(0,0)$}. 
I am just going to provide the first function f, as I just want to get an idea of how to do this, and hopefully I can do the rest on my own.
i) $f(x,y) := \frac{cos(x² + y²)}{x² + y²}$
So the first thing I thought if, is that function is defined everywhere except for (0,0). Well I thought, what happens when we let the $(x,y)$ approach $(0,0)$, the function is going to tend to infinity.
If I understood the question correctly, I think I need to find a new function , if it exists, that has the exact same behavior of $f$, but that is also defined at $(0,0)$, where if I let $(x,y)$ tend to $(0,0)$ the limit should be $\tilde{f}(0,0) = \infty$. 
This cannot be guessing work, so I am sure there must be a mathematical way to approach this, and that is where I hope you guys could guide me in the right direction.
Thanks!
EDIT:
The other functions are: 
$$g(x,y) := \frac{p(x^2+y^2)}{x^2+y^2}$$ where $p: \mathbb{R} \rightarrow \mathbb{R}$ is a continuous function with $|p(r)| \leq 2|r|^2$
and 
$$h(x,y) := \frac{x^2+y^2}{|x-y|^2+x^4+y^4}$$
I am really confused by this question and still not sure how to solve it. Further help would be appreciated.
 A: There is multiple ways to answer the question. 
The key is to compute $\underset{(x,y)\to (0,0)}\lim f(x,y)$.
If the limit exists and is finite, then you can extend the function $f$ to a continuous function on the whole plane. Then it is natural to define :
$$\tilde f(x,y)=\left\{\begin{matrix}
f(x,y) & \text{if } (x,y)\neq(0,0)\\ 
 l& \text{if } (x,y)=(0,0)
\end{matrix}\right.$$
Where $l=\underset{(x,y)\to (0,0)}\lim f(x,y)$.
Here, you may just set $y=0$ and compute $$\underset{x \to 0}\lim f(x,0)= \underset{x \to 0}\lim\frac {cos(x^2)} {x^2}=\infty.$$
Hence, even if $\underset{(x,y)\to (0,0)}\lim f(x,y)$ exists, it will not be finite, so the function $\tilde f$ does not exist.
EDIT :
For $g$ and $h$, try using polar coordinates.
Set $x=r \cos( \theta)$ and $y=r \sin( \theta)$. Then use the fact that, if the limit exists, then :
$$\underset{(x,y)\to (0,0)}\lim f(x,y)=\underset{r\to 0}\lim f(r cos(\theta),r \sin (\theta)).$$
EDIT 2 :
$g(r \cos(\theta),r \sin(\theta))=\frac {p(r^2)}{r^2}$.
Since $|p(r)| \leq 2 |r^2|$, we have :
$$|g(r \cos(\theta),r \sin(\theta))|\leq 2 \frac {|r^4|} {|r^2|}=2 r^2 \underset{r\to 0}\to 0.$$
Then we can conclude that $g$ can be extended on the whole plane by $\tilde g$ with $\tilde g (0)=0$.
