An intuitive doubt on limits

Question

So I have been studying iterated limits and I have a doubt.
The iterated limits of a two variable function $f(x,y) \mbox{ at } (\alpha, \beta)$ are defined as : $$\lim_{x \to \alpha} (\lim_{y \to \beta} f(x,y)) \mbox{ and } \lim_{y \to \beta} (\lim_{x \to \alpha} f(x,y)) $$ The thing is that the iterated limit exists only if the internal one related to it exists, so : $$ \lim_{x \to \alpha} (\lim_{y \to \beta} f(x,y)) \mbox{ exist iff } (\lim_{y \to \beta} f(x,y)) \mbox{ exist independently } $$ same is true for the other iterated limit.
now look at this function : $$f(x,y) = xsin(1/y)$$ the limit: $$\lim_{y \to 0} xsin(1/y) \mbox{ does not exist}$$ and that makes sense as y tending to zero causes $sin(1/y)$ to oscillate and it will be some undetermined but finite number(changing rapidly as y approaches zero) times $x$ at the end, which means we don't know the limit.
But here in reality, later $x$ also approaches zero, which means that at the end its : $$\mbox{(some undetermined but finite number, oscillating rapidly)} \times \mbox{ (a number approaching 0)}$$ logically, no matter what undetermined value $sin(1/y)$ takes, that value will be multiplied to a number really close to zero, so the limit should be zero. But this is not the case.
According to various references and my professor the limit : $$ \lim_{x \to 0} (\lim_{y \to 0} xsin(1/y)) \mbox{ does not exist}$$ This seems paradoxical and counterintuitive , please share your wisdom and point where I went wrong, or if this is right.

Answer 1

I am not sure if I fully understand your question. To define a limit (iterated in this case), or any other concept for that matter, in a well defined manner one must make sure that the values of limits make sense and are uniquely defined in ALL cases. Just because in the special case of x sin(1/y) we can argue that the limit exists does not warrant the general definition to allow for cases where the independent limit is not defined.

In other words, the definition does not imply that the limit CAN NEVER BE DEFINED unambiguously. Only that, if you need it to be defined unambiguously regardless, then you should assume $ (\lim_{y \to \beta} f(x,y))$ exist independently. As a matter of fact, you might be able to weaken this assumption (intuitively, not by much). You might be able to show that in some cases $\lim_{x \to \alpha} (\lim_{y \to \beta} f(x,y)) \mbox{ exist}$ even if $(\lim_{y \to \beta} f(x,y))$ does not exist independently.

Definitions are designed to work. They are not necessarily optimal.

Answer 2

About your example $$x\sin \frac 1y,$$ the limit does not exist at $(0,0)$ because if you fix any value of $x$ (you may take $x=1$ or whatever), you find that the limit as $y\to 0$ is undefined. On the other hand fix a value of $y\ne 0$ (say take $y=1/π$). Then the limit as $x\to 0$ is $0.$ Since for the limit to exist it shouldn't matter how $(x,y)$ approaches the origin, it follows that the limit doesn't exist.

Your mistake is taking the limit of an undefined quantity. Yes, the quantity is bounded, but that doesn't tell us exactly what it is. So taking the outer limit is quite meaningless -- it is undefined as well. Or, in other words, does not exist.