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$f(x,y)= \begin{cases} x \sin{\frac{1}{y}}, & \text{$y \ne 0$} \\ 0, & \text{$y=0$} \end{cases}$

Study the following limits: $$\lim _{x \rightarrow 0} {f(x,y)}$$ $$\lim _{y \rightarrow 0} {f(x,y)}$$ $$\lim _{(x,y) \rightarrow (0,0)} {f(x,y)}$$

The limit $\lim _{x \rightarrow 0} {f(x,y)}=0$ (as it is $0$ on both cases), and the limit $\lim _{y \rightarrow 0} f(x,y)$ doesn't exist.

However, doing an $\epsilon - \delta$ proof for the limit $\lim _{(x,y) \rightarrow (0,0)} {f(x,y)}$ shows that it exists and it is equal to $0$.

My question is, since the limit $\lim _{y \rightarrow 0} {f(x,y)}$ doesn't exist, shouldn't this be an indicator that the limit $\lim _{(x,y) \rightarrow (0,0)} {f(x,y)}$ doesn't exist?

(Are there any other examples for similarly tricky situations?)

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    $\begingroup$ Squeeze.. $$-1 \le \sin(1/y) \le 1 \implies -x \le x \sin(1/y) \le x$$ Now what can you say? Also, by your definition of $f$ in the question, the limit as $y \to 0$ does exist. $\endgroup$ Oct 11, 2017 at 8:23
  • $\begingroup$ As @Mattos wrote, the $sin$ function is bounded and $x\rightarrow 0$... $\endgroup$
    – gbox
    Oct 11, 2017 at 8:28
  • $\begingroup$ What should be a indicator that the limit doesn't exist is the limit $\lim_{y\to 0} f(0,y)$ (fixing $x=0$ because path has to pass through $(0,0)$), and it is equal $0$. $\endgroup$ Oct 11, 2017 at 8:31

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and the limit $\lim _{y \rightarrow 0} f(x,y)$ doesn't exist.

Wrong.

The correct version would be:

If $x\neq 0$, the limit $\lim_{y\to0} f(x,y)$ does not exist.

For $x=0$, the limit is equal to $$\lim_{y\to 0} 0\cdot\sin\frac1y$$

and this limit is clearly equal to $0$.

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  • $\begingroup$ Oops, my bad. But doesn't the fact that the limit $\lim _{y \rightarrow 0} {f(x,y)}$ doesn't exist, automatically make the $\lim _{(x,y) \rightarrow (0,0)} {f(x,y)}$ not exist? Aren't we approaching the origin from a direction in that way? $\endgroup$ Oct 11, 2017 at 9:20
  • $\begingroup$ @zeropancakes that would only be true if the limit did not exist for y=0 $\endgroup$
    – 5xum
    Oct 11, 2017 at 9:24
  • $\begingroup$ Did you mean for x=0? $\endgroup$ Oct 11, 2017 at 9:27
  • $\begingroup$ @zeropancakes yes, of course. Sorry. $\endgroup$
    – 5xum
    Oct 11, 2017 at 9:27
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    $\begingroup$ But how is it not continuous? Don't we have $\lim _{x \rightarrow 5} {f(x,0)}=\lim _{x \rightarrow 5} {0}=0=f(5,0)$ ? $\endgroup$ Oct 11, 2017 at 9:39

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