If \(\lim _{x \rightarrow x_{0}} y(x)=y_{0}\) and \(\lim _{x \rightarrow x_{0}}
f(x, y(x))=L,\) we say that \(f(x, y) a p-\)
proaches \(L\) as \((x, y)\) approaches \(\left(x_{0}, y_{0}\right)\) along the
curve \(y=y(x)\).
(a) Prove: If \(\lim _{(x, y) \rightarrow\left(x_{0}, y_{0}\right)} f(x, y)=L,\)
then \(f(x, y)\) approaches \(L\) as \((x, y)\) approaches \(\left(x_{0},
y_{0}\right)\) along any curve \(y=y(x)\) through \(\left(x_{0}, y_{0}\right)\)
(b) We saw in Example 5.2 .3 that if
$$
f(x, y)=\frac{x y}{x^{2}+y^{2}},
$$
then \(\lim _{(x, y) \rightarrow(0,0)} f(x, y)\) does not exist. Show, however,
that \(f(x, y)\) approaches a value \(L_{a}\) as \((x, y)\) approaches (0,0) along
any curve \(y=y(x)\) that passes through (0,0) with slope \(a\). Find \(L_{a}\).
(c) Show that the function
$$
g(x, y)=\frac{x^{3} y^{4}}{\left(x^{2}+y^{6}\right)^{3}}
$$
approaches 0 as \((x, y)\) approaches (0,0) along a curve as described in (b),
but that \(\lim _{(x, y) \rightarrow(0,0)} f(x, y)\) does not exist.