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.

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