Show that $\lim\limits_{(x, y)\rightarrow (0, 0)}\dfrac{xy}{ x^{2}+y^{2}}$ does not exist.

Solution  Look at the denominator. Since $\lim\limits_{(x, y) \rightarrow (0, 0)}( x^{2}+y^{2}) =0$, we cannot use the limit of a quotient to investigate the limit. We investigate the limit using two curves that contain $(0,0)$, say, $y=2x$ and $y=-x.$

First we investigate the limit using the line $y=2x$. Then $$\begin{equation*} {\rm Using}\; y=2x: \quad \lim\limits_{(x, y)\rightarrow (0, 0)}\dfrac{x\left( 2x\right) }{x^{2}+\left( 2x\right) ^{2}}=\lim\limits_{x\rightarrow 0}\dfrac{ 2x^{2}}{x^{2}+4x^{2}}=\lim\limits_{x\rightarrow 0}\dfrac{2x^{2}}{5x^{2}}= \dfrac{2}{5} \end{equation*}$$

Figure 23 $f(x,y) =\dfrac{xy}{x^{2}+y^{2}}$

Now we investigate the limit using the line $y=-x$. $$\begin{equation*} {\rm Using}\; y=-x: \quad \lim\limits_{(x, y)\rightarrow (0, 0)}\dfrac{x\left( -x\right) }{x^{2}+\left( -x\right) ^{2}}=\lim\limits_{x\rightarrow 0}\dfrac{ -x^{2}}{x^{2}+x^{2}}=\lim\limits_{x\rightarrow 0}\dfrac{-x^{2}}{2x^{2}}=- \dfrac{1}{2} \end{equation*}$$

Since two different answers are obtained, the limit does not exist.