Prove that the directional derivatives of
$$
f(x, y)=\left\{\begin{array}{ll}
\frac{x^{2} y}{x^{4}+y^{2}} & (x, y) \neq(0,0) \\
0 & (x, y)=(0,0)
\end{array}\right.
$$
exist at $(0,0)$ in all directions $\boldsymbol{u},$ but $f$ is neither continuous nor differentiable
at $(0,0)$
Proof. Let $(u, v)$ be a unit vector. By definition,
$$
D_{(u, v)} f(0,0)=\lim _{t \rightarrow 0} \frac{f(t u, t v)-f(0,0)}{t}=\lim _{t \rightarrow 0} \frac{u^{2} v}{u^{4} t^{2}+v^{2}}=\left\{\begin{array}{lr}
u^{2} / v & v \neq 0 \\
0 & v=0
\end{array}\right.
$$
Thus the directional derivatives of $f$ exist. On the other hand, $f$ is not continuous because along the path $x=0$ the limit is zero, but along the path $y=x^{2}$ the limit is $1 / 2$
My question is Why did we say that "if $v\neq 0$ and $v=0 $" in the above piecewise function, I couldn't understand. Where is this coming from? Can you explain it? Why didn't we say $(u,v)\neq (0,0)$ or $(u,v)=(0,0)$ above piecewise function? Thanks…
Best Answer
$$ \frac{u^{2} v}{u^{4} t^{2}+v^{2}} \to \frac{u^{2} v}{v^{2}}= \frac{u^{2}}{ v}$$
as $t \to 0.$