I need to prove that $f$ continuous at $(x)=0$ using a $\epsilon$-$\delta$ proof I need to prove that $f$  continuous at $(x)=0$ using a $\epsilon$-$\delta$ proof
$$
  f(x, y) = \begin{cases}
x^2sin(\frac1x),&x\neq 0
\\
0,&x = 0
\end{cases}
$$
 A: If $0<|x|<\delta$ then $$|f(x)-f(0)| = |x^2 \sin(1/x)|\le|x|^2 < \delta^2$$
Thus we for every $\epsilon>0$ we can choose $\delta = \sqrt{\epsilon}$ such that... 
A: Hint: For $x\neq 0$, $$|f(x)-f(0)|=\left|x^2\cdot\sin\frac1x-0\right|=|x|^2\left|\sin\frac1x\right|\leq|x|^2=|x-0|^2.$$
A: This post is only meant to be a supplement to the answers posted above.(So that the process is understood.)
You have to prove that the limit of the function as x tends to zero is the same as the value of the function at zero.
Here epsilon is the upper bound on the distance between an abitrary f(x) and and f(0). And delta is the upper bound on the distance between an arbitrary x and 0. If you can prove that you can bring f(x) as close to f(0) as you wish, by bringing x as near to zero, you prove that the limit of the function as x tends to zero exists.(This can be done without the epsilon - delta definition, by using algebraic manipulation, but it won't be rigorous.More rigorously, for any arbitrary epsilon, you should be able to find a delta - and for this you have to use the epsilon-delta definition.) 
Note that epsilon and delta are both greater than zero, so this excludes the case f(x) = f(0) and the case x = 0. So it only proves that the limit exists, it says nothing about the behaviour of the function at x= 0. So you have to prove that the limit above is equal to the value of f(x) at x = 0. Here since it is already given that f(0) = 0, all you have to do is to state the fact.(A practically trivial, but logically important step).
