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Show, with the definition, that $\lim_\limits{ (x,y) \to (0,0)} x\sin\frac{1}{y} + y\sin\frac{1}{x} $ exist; $(x,y) \in \mathbb{R^2}-\{(0,0)\}$.

I think the limit is zero because for $||(x,y)|| < \delta$, $$||x\sin\frac{1}{y} + y\sin\frac{1}{x}|| \leq ||(x,y)|| \cdot||(\sin\frac{1}{y},\sin\frac{1}{x})|| \leq \sqrt{2} ||(x,y)||< \sqrt{2}\delta.$$ It is sufficient to define that $\sqrt{2}\delta = \epsilon$

I am not certain of what I did so far. Is there anyone who can give me a hint to solve the problem?

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  • $\begingroup$ You're basically done. Can you state the definition of limit and mention exactly what keeps you from finishing? $\endgroup$ – Git Gud Jan 30 '16 at 17:24
  • $\begingroup$ You need that $x,y\neq 0$. It is not enough to exclude just the point $(0,0)$ - this function is still undefined if $x=0$ and $y\neq 0$. $\endgroup$ – Thomas Andrews Jan 30 '16 at 17:25
  • $\begingroup$ Just tell me if that $..$ means it is done or do you think there is some thing more than just $||(x,y)$|| $\endgroup$ – kk lm Jan 30 '16 at 17:26
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    $\begingroup$ The first inequality is Cauchy-Schwartz. The second is wrong as it is ($x=y=\dfrac{1}{2n\pi+\frac\pi2}$), but it is true if you substitute $\lVert (x,y)\rVert$ with $\sqrt{2}\lVert (x,y)\rVert$. $\endgroup$ – user228113 Jan 30 '16 at 17:27
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    $\begingroup$ Full proof that the limit exists and that it is $0$: $$\left|x\sin\frac{1}{y} + y\sin\frac{1}{x}\right|\leqslant|x|+|y|$$ $\endgroup$ – Did Jan 30 '16 at 17:27
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You guessed correctly and that was crucial to make the next step. Let $\varepsilon>0$ and we shall find $\delta>0$ such that for all $(x,y)$ on the domain of $f$ such that $\|(x,y)-(0,0)\|_2<\delta$ we get $|f(x,y)-0|<\varepsilon$ (here $\|\cdot\|_2$ denotes the Euclidean metric). Indeed, if $\delta=\varepsilon/2>0$ then for all $(x,y)$ on the domain of $f$ such that $\|(x,y)-(0,0)\|_2<\delta$ we get: $$|f(x,y)-0|\leq |x||\sin(1/y)|+|y||\sin(1/x)|\leq |x|+|y|\leq \|(x,y\|_2+\|(x,y)\|_2<2\delta=\varepsilon.$$

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