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Use the definition of a limit to show that:

$$\lim_{(x,y)\to (0,0)} (x+y)\sin\left(\frac{1}{x}\right)\cos\left(\frac{1}{x}\right) = 0$$

Id appreciate the thought process behind the solution, the epsilon-delta definition never makes sense in my head.

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    $\begingroup$ Hint: Recall that $\sin$ and $\cos$ are bounded by $1$, so you should squeeze this. $\endgroup$ – Michael Burr Mar 2 '17 at 14:29
  • $\begingroup$ The "squeeze" technique is shown in the A by Menchun Zhang. Find a (simpler) upper bound $g(x,y)$ for $|f(x,y)|$ and show that $g(x,y)\to 0$. That implies |$f(x,y)|\to 0$. $\endgroup$ – DanielWainfleet Mar 2 '17 at 16:38
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Here is the thought process of how to work out a $\,\delta\,$

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For all $\,\varepsilon>0$, we need to find a $\,\delta\,$ such that for all $\,(x,y)\,$ with

$$\sqrt{x^2+y^2}<\delta$$

we have

$$\left|\,(x+y)\sin\left(\frac1x\right)\cos\left(\frac1x\right)\,\right|<\varepsilon$$

To find such a $\,\delta$, we need to insert something between $\,\varepsilon\,$ and the left absolute value in the above inequality. Observe that $\,|\sin(1/x)|\leq1,|\cos(1/x)|\leq1\,$, so we could insert $\,|x+y|$ as below

$$\left|\,(x+y)\sin\left(\frac1x\right)\cos\left(\frac1x\right)\,\right|\leq|x+y|<\varepsilon$$

Next, since $\,|x+y|\leq|x|+|y|\leq2\sqrt{x^2+y^2}$, we could insert $\,2\sqrt{x^2+y^2}\,$ again as below

$$|x+y|\leq2\sqrt{x^2+y^2}<\varepsilon$$

Now because $\,\sqrt{x^2+y^2}<\delta\,$, we finally insert $\,\delta\,$ in the inequality,

$$2\sqrt{x^2+y^2}<2\delta\leq\varepsilon$$

Thus, here is $\,0<\delta\leq\varepsilon/2\,$ and we can simply let

$$\delta=\frac\varepsilon2$$

And that is the answer


This is only the thought, so you need to reverse the whole process to get a formal answer, and that has been nicely done by Gensan

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  • $\begingroup$ Thank you for the reasoning behind the answer. This is very helpful. $\endgroup$ – Joseph Moraine Mar 2 '17 at 16:00
  • $\begingroup$ @JosephMoraine No worries, it is good to hear that helped you. $\endgroup$ – Mengchun Zhang Mar 2 '17 at 16:40
  • $\begingroup$ It's a delta not an delta $\endgroup$ – Displayname Mar 2 '17 at 17:48
  • $\begingroup$ Oops, no idea why I made that mistake, but thanks anyway! $\endgroup$ – Mengchun Zhang Mar 2 '17 at 17:54
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Let $\epsilon>0$. Take $\delta=\frac{\epsilon}{2}$. Let $(x,y)\in\Bbb R^2$ such that $0<|(x,y)-(0,0)|<\delta$. This means that $$0<\sqrt{(x-0)^2+(y-0)^2}<\delta.$$ We get $$x^2\leq x^2+y^2<\delta^2\quad\text{and }\quad y^2\leq x^2+y^2<\delta^2.$$ Thus, $$|x|<\delta\quad\text{and}\quad |y|<\delta.$$ Therefore, $$\begin{align} \bigg|\bigg[(x+y)\sin\frac{1}{x}\cos\frac{1}{x}\bigg]-0\bigg|&=|x+y|\cdot\bigg|\sin\frac{1}{x}\bigg|\cdot\bigg|\cos\frac{1}{x}\bigg|\\ &\leq |x+y|\\ &\leq |x|+|y|\\ &<\delta+\delta=2\delta=\epsilon. \end{align} $$ Hence, $$\lim_{(x,y)\to (0,0)} (x+y)\sin\left(\frac{1}{x}\right)\cos\left(\frac{1}{x}\right) = 0.$$

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