Use the $\varepsilon$-$\delta$ definition of a limit to show that $\lim\limits_{(x, y)\rightarrow (0, 0)}\dfrac{2xy^{2}}{x^{2}+y^{2}}=0.$

Solution  Given a number $\varepsilon >0$, we seek a number $\delta >0$, so that $$\begin{equation*} \hbox{whenever}\quad 0<\sqrt{(x-0) ^{2}+( y-0) ^{2}} <\delta \qquad \hbox{then}\quad \left\vert \dfrac{2xy^{2}}{x^{2}+y^{2}} -0\right\vert <\varepsilon \end{equation*}$$

That is, $$\begin{equation*} \hbox{whenever }\quad 0<\sqrt{x^{2}+y^{2}}<\delta \quad \hbox{then }\quad \left\vert \dfrac{2xy^{2}}{x^{2}+y^{2}}\right\vert <\varepsilon \end{equation*}$$

We need to find a connection between $\dfrac{2xy^{2}}{x^{2}+y^{2}}$ and $\sqrt{x^{2}+y^{2}}.$ We begin with the observation that since $x^{2}\geq 0,$ $$\begin{equation*} y^{2}\leq x^{2}+y^{2}\hbox{ } \end{equation*}$$

So, for all points $(x,y)$ not equal to $(0,0),$ we have $$\dfrac{y^{2}}{x^{2}+y^{2}}\leq 1\qquad {\color{#0066A7}{\hbox{Divide both sides by \(x^2+y^2\).}}}$$

Now $$\begin{equation*} \left\vert \frac{2xy^{2}}{x^{2}+y^{2}}\right\vert =\frac{2\vert x\vert y^{2}}{x^{2}+y^{2}}=2\vert x\vert \cdot \dfrac{ y^{2}}{x^{2}+y^{2}}\leq 2\left\vert x\right\vert \cdot 1=2\sqrt{x^{2}}\leq 2 \sqrt{x^{2}+y^{2}} \end{equation*}$$

Given $\varepsilon >0$, we choose $\delta \leq \dfrac{\varepsilon }{2}$. Then whenever $0<\sqrt{x^{2}+y^{2}}<\delta$, we have $$\begin{equation*} \left\vert \frac{2xy^{2}}{x^{2}+y^{2}}\right\vert \leq 2\sqrt{x^{2}+y^{2}} <2\delta \leq \varepsilon \end{equation*}$$

That is, $$\lim_{(x, y)\rightarrow (0, 0)}\left\vert \frac{2xy^{2}}{x^{2}+y^{2}} \right\vert =0 %%\tag{$\blacksquare \(}$$