Extending Function to Make it Continuous I've been struggling with extending this function to make it continuous:
$$f(x,y)=\frac{1}{x}\sin\left(\frac{x^3}{x^2+y^2}\right)$$
So, the domain is $\mathbb{R}^2\backslash\{(0,y):y\in\mathbb{R}\}$.
Set of limit points of the domain is obviously $\mathbb{R}^2$.
I'm not sure how to approach this problem because none of the methods we've done in class work.
I'm not even sure how to a priori assume whether I can or can't extend this function to be continuous (or even in which points is it possible to do so).
This is all new to me so I would appreciate any help or advice on how to predict the behaviour of this function.
Thanks in advance.
 A: This function cannot be extended continuously to the entire plane $\mathbb{R}^2$, because there are points on the line $x=0$ such that your function $f$ approaches different limits from different directions.
For example, consider the point $P=(0,0)$, which does not lie in the domain of $f$, and consider the limit of $f(x,y)$ as $(x,y)$ approaches this point along the two lines $y=0$ and $y=x$. Clearly $f$ is defined at all points on these lines except $P$.
On the line $y=0$, $f(x,y)=f(x,0)=\frac{1}{x}sin(\frac{x^3}{x^2+0})=\frac{1}{x}sin(x)=\frac{sin(x)}{x}$. 
So using l'Hopitals rule, we see that $lim_{x\rightarrow 0}\frac{sin(x)}{x}=lim_{x\rightarrow 0}\frac{cos(x)}{1}=cos(0)=1$.
Similarly, on the line $y=x$, $f(x,y)=f(x,x)=\frac{1}{x}sin(\frac{x^3}{x^2+x^2})=\frac{1}{x}sin(\frac{1}{2}x)$, 
and again, using l'Hopital's rule, $lim_{x\rightarrow 0}\frac{sin(\frac{1}{2}x)}{x}=lim_{x\rightarrow 0}\frac{\frac{1}{2}cos(\frac{1}{2}x)}{1}=\frac{1}{2}cos(0)=\frac{1}{2}$.
This proves that the limit of $f$ as $(x,y)$ approaches $P$ is undefined, and so $f$ cannot be extended continuously to the whole plane.
A: Going to detail my ``elsewhere'' in the previous comment. Fix $y\neq 0$, then for $x<<1$ the argument of the sine is small thus it is comparable to the sine itself plus lower order terms $o(x)$, i.e.
$$\frac{1}{x}\sin\left(\frac{x^3}{x^2+y^2}\right) \approx \frac{x^2}{x^2+y^2} +o(1)$$
As you send $x\to 0$ you get that it has as limit $0$, hence you can continuously extend to $(0,y)$ for all $y\neq 0$ by setting
$$f(x,y) = 0\,, \qquad \text{on $\{(0,y)\,:\, y\neq 0\}$}$$
As Guy Fasone example shows, at the origin this trick does not work. 

By the way, the set of limit points at the origin is $[0,1]$, not the whole $\mathbb{R}$. Passing to polar coordinates you get
$$\frac{1}{\rho \cos \theta} \sin\left(\rho \cos^3 \theta \right) \approx cos^2 \theta$$
A: 
You can either extend $f$ as follow 
  $$f(x,y) =\begin{cases}\displaystyle\frac{1}{x}\sin\left(\frac{x^3}{x^2+y^2}\right)& x\neq 0\\0&x=0 \end{cases}$$ or 
  $$f(x,y) =\begin{cases}\frac{1}{x}\displaystyle\sin\left(\frac{x^3}{x^2+y^2}\right)& x\neq 0\\1&x=0 \end{cases}$$ 
  Remark: The first extension is continuous everywhere except at $(0,0)$ (see below) whereas the second is not continuous on $\{(0,y)\in\Bbb R^2\}.$

However, The function $f$ will never be continuously extended  at $(0,0)$
since, 
For all $h\neq 0$ and we have, $$(h,0), (h,y)\in \mathbb{R}^2\backslash\{(0,y):y\in\mathbb{R}\}$$
but, 
$$\lim_{h\to 0}f(x,0) =\lim_{h\to 0}\frac{\sin h}{h}= 1$$
and  for $y \neq 0$ near $h$ near zero we have,
 $$f(h,y) =\frac{1}{h}\sin\left(\frac{h^3}{h^2+y^2}\right) \sim\frac{h^2}{h^2+y^2}\to 0\neq 1$$

Continuity of the first extension on $\Bbb R^2\setminus\{(0,0)\}$
  This last line also prove that if $y_0\neq 0$ then 
  $$\lim_{(x,y)\to (0,y_0)}f(x,y) = 0 = f(0,y_0)$$
  which end the proof since naturally $f$ is continuous on its domain $\mathbb{R}^2\backslash\{(0,y):y\in\mathbb{R}\}$

