Showing That a Limit Does Not Exist

Show that \(\lim\limits_{(x, y)\rightarrow (0, 0)}\dfrac{x^{2}y}{x^{4}+y^{2}}\) does not exist.

Solution  We investigate the limit using curves that contain \( (0,0).\) Based on our experience with the function in Example 2, we can simultaneously test all nonvertical lines that contain \((0,0)\) by using \( y=mx \). Then \[ \begin{equation*} {\rm Using}\; y=mx: \quad \lim\limits_{(x, y)\rightarrow (0, 0)}\dfrac{x^{2}\left( mx\right) }{x^{4}+\left( mx\right) ^{2}}=\lim\limits_{x\rightarrow 0}\dfrac{ mx^{3}}{x^{4}+m^{2}x^{2}}=\lim\limits_{x\rightarrow 0}\dfrac{mx}{x^{2}+m^{2}} =0 \end{equation*} \]

Even if the same limit is obtained along two curves that contain \(P_{0}\) (or three, or even several hundred curves), we cannot conclude that the limit of the function exists. For instance, in Example 2, using the curves \(y=2x\) and \(y=\dfrac{1}{2}x\) would both result in the same value for the limit, but we have shown the limit does not exist.

Along every nonvertical line containing the point \((0,0)\), the limit is \(0\). But this does not mean that \(\lim\limits_{(x, y)\rightarrow (0, 0)}\dfrac{ x^{2}y}{x^{4}+y^{2}}\) is \(0\). Suppose we use the curve \(y=x^{2}\). Then \[ \begin{equation*} {\rm Using}\; y=x^{2}: \quad \lim\limits_{(x, y)\rightarrow (0, 0)}\dfrac{x^{4}}{ x^{4}+x^{4}}=\lim\limits_{x\rightarrow 0}\dfrac{x^{4}}{2x^{4}}=\dfrac{1 }{2} \end{equation*} \]

Since two different curves that contain \((0,0) \) result in different values for the limit, the limit does not exist.