Q:

Prove that $\lim_{y\to0}\lim_{x\to0}(x+y\sin\frac{1}{x}) $ doesn't exist

Actually the full problem states: given a function $f(x, y) = x + y\sin\frac{1}{x}$ prove that $\lim_{x\to0}\lim_{y\to0}f(x, y)$ and $\lim_{(x, y) \to(0,0)}f(x, y) $ exist and that $\lim_{y\to0}\lim_{x\to0}f(x, y)$ doesn't exist.
The first two weren't a problem really, first one is just direct calculation the other one is applying the three limits theorem. The third one confuses me. I remember there being a theorem that states something like the following:

$\lim_{x \to x_0}f(x) = A$ if and only if for every sequence of real
  numbers $(x_n)$ where $\forall n \quad x_n \neq x_0$ such that
  $\lim_{n \to \infty}x_n = x_0$, then $\lim_{n \to \infty}f(x_n) = A$

However, in the textbook they seem to be using some other property that they haven't mentioned so far. I'm quoting now:

We will prove that $\lim_{x->0}f(x, y)$ cannot exist, therefore the
  entire limit cannot exist. Let's look at the sequences
  $a_n=(\frac{1}{n\pi})$ and $b_n =(\frac{1}{2n\pi + \frac{\pi}{2}})$.
  $$\lim_{n \to
> \infty}f(a_n, y) = \lim_{n \to \infty}(\frac{1}{n\pi} +0\cdot y) = 0$$
$$\lim_{n \to \infty}f(b_n, y) = y$$
Therefore the limit cannot exist.

So it kind of looks like some application of the previous theorem? But still then, it looks different because the theorem doesn't state anything about the existence of the limit as far as I understood. Is this some separate theorem that I've missed?
Thanks.

A:

A corollary of the previously mentioned theorem is that

if you can find two different sequences $(a_n),(b_n)$ converging to $x_0$, but for which $(f(a_n))$ and $(f(b_n))$ converge to different limits, then $\lim_{x\to x_0}f(x)$ does not exist.

Thus the demonstration that $\lim_{y\to0}\lim_{x\to0}f(x,y)$ does not exist makes use of the theorem. It finds two sequences that converge to the same limit by themselves ($0$), but different limits once the function is applied. $y$ need not be $0$, so the limit is undefined if $y\ne0$, and the whole limit does not exist.

