In this problem we show that the function

\begin{equation*}
f(x,y) = \frac{2x-y^{2}}{x+y^{2}}
\end{equation*}

does not have a limit as \((x,y)\to (0,0)\text{.}\)

*(a)* Suppose that we consider \((x,y)\to (0,0)\) along the curve \(y = 4 x^{1/2}\text{.}\) Find the limit in this case:

\(\lim\limits_{(x,4 x^{1/2})\to(0,0)} \frac{2x-y^{2}}{x+y^{2}} =\)

*(b)* Now consider \((x,y)\to (0,0)\) along the curve \(y = 5 x^{1/2}\text{.}\) Find the limit in this case:

\(\lim\limits_{(x,5 x^{1/2})\to(0,0)} \frac{2x-y^{2}}{x+y^{2}} =\)

*(c)* Note that the results from *(a)* and *(b)* indicate that \(f\) has no limit as \((x,y)\to (0,0)\) *(be sure you can explain why!)*.

To show this more generally, consider \((x,y)\to (0,0)\) along the curve \(y = m x^{1/2}\text{,}\) for arbitrary \(m\text{.}\) Find the limit in this case:

\(\lim\limits_{(x,m x^{1/2})\to(0,0)} \frac{2x-y^{2}}{x+y^{2}} =\)

*(Be sure that you can explain how this result also indicates that \(f\) has no limit as \((x,y)\to(0,0)\text{.}\)*