Showing That a Limit Exists

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 \(} \]