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Show that $f(x,y) = \frac{\sin(xy)}{|x-y|}$ has no limit when $(x,y) \to (0,0)$. Hint: try subsequences.

We worked part of this out in class and I'm going over it to arrive at a final solution and I'm somewhat stuck.

The first thing we did was reduce it to a one dimensional problem by letting $$g(x,mx) = \frac{\sin(mx^{2})}{|x-mx|} = \frac{\sin(mx^{2})}{|x||1-m|}$$

Which we can manipulate:

$$\lim_{x \to 0}g(x,mx) = \lim_{x \to 0} \frac{\sin(mx^{2})}{|x||1-m|} \frac{mx^{2}}{mx^{2}} = 1 \cdot 0 = 0$$

So we showed along one path that the function goes to zero. The hint now says to show by subsequences that this function does not exist. To do that I would have to find a convergent sequence, $a_{n} \to 0$ such that the function $g(a_{n}) \nrightarrow 0$. From this I could then pick an epsilon and prove it by contradiction.

My question is how can I find this subsequence?

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We have that

$$\frac{\sin(xy)}{|x-y|}=\frac{\sin(xy)}{xy}\frac{xy}{|x-y|}$$

and $\frac{\sin(xy)}{xy}\to 1$ but for $x=2y$

$$\frac{xy}{|x-y|}=\frac{2y^2}{|y|} \to 0$$

and for $x=t+t^2$ and $y=t\to 0$

$$\frac{xy}{|x-y|}=\frac{t^2+t^3}{t^2} \to 1$$

therefore the given limit doesn't exist.

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$x_{n}=\dfrac{n}{(n+1)^{2}}$, $y_{n}=\dfrac{1}{n}$, $\sin(x_{n}y_{n})=\sin\dfrac{1}{(n+1)^{2}}$, and $|x_{n}-y_{n}|=\dfrac{1}{(n+1)^{2}}\left|-2-\dfrac{1}{n}\right|$, the limit as quotient goes to $1/2$.

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