Using the Epsilon-Delta definition of a limit 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.
 A: 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,
$$\lim_{(x,y)\to (0,0)} (x+y)\sin\left(\frac{1}{x}\right)\cos\left(\frac{1}{x}\right) = 0.$$
A: Here is the thought process of how to work out a $\,\delta\,$
$\left.\right.$
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
