Suppose a function f: $R^2 \rightarrow R$ is defined by $\frac{xy^2}{x^2+y^4}$ if x,y is non zero, and f(x,y) = 0 if (x,y)=(0,0). Show that for all unit vectors u the directional derivative of f at the origin in the direction u does exist, but f is discontinuous at (0, 0). Show that there is no plane which contains all the lines which are tangent to the surface z = f(x, y) at (0, 0, 0).
Comments: 1. For the first part, I am approaching this by trying to prove that the function satisfies the definition of differentiability, namely that there exists a linear map L such that $\frac{||f(x)-f(0,0)-L(x-(0,0))||}{||x||}$ as x->0 is 0. So I got that the Jacobian was a zero matrix, so we just get $\frac{xy^2}{(x^2+y^4) \sqrt{x^2+y^2}}$ but I don't think this goes to zero? Have I made a mistake somewhere? I am aware that there are other ways to prove differentiability (such as using the fact that if partials exist and are continuous), but why is this way not getting me the answer?
- Could someone help me with the part about showing there's no plane? Would it be along the lines of showing that there's no single normal vector common to all the derivative vectors at (0,0)?
Let’s start with continuity. For $f$ to be continuous at the origin, its limit along every path to the origin must be equal to its value there. Consider the curve $x=y^2$: $$\lim_{y\to0}f(y^2,y) = \lim_{y\to0}{y^4\over2y^4} = \frac12 \ne f(0,0).$$ Since $f$ is not continuous at the origin, it’s not differentiable, either.
On the other hand, let $u=(\cos\theta,\sin\theta)$ for fixed $\theta$. The derivative in the direction of $u$ is $$\begin{align} \lim_{h\to0^+}\frac1h \left[f(h\cos\theta,h\sin\theta)-f(0,0)\right] &= \lim_{h\to0^+}\frac1h{(h\cos\theta)(h\sin\theta)^2 \over (h\cos\theta)^2+(h\sin\theta)^4} \\ &= \lim_{h\to0^+}{\cos\theta \sin^2\theta \over \cos^2\theta + h^2\sin^4\theta} \\ &= {\sin^2\theta\over\cos\theta} \end{align}$$ if $\cos\theta\ne0$, which corresponds to derivatives in the direction of the $y$-axis. For those cases $f(0,y)=0$, so the directional derivative, which is just $f_y(0,0)$, is equal to $0$. We also have $f(x,0)=0$, so $f_x(0,0)=0$, therefore if there were a tangent plane, it would have to be the $x$-$y$ plane. This means that all directional derivatives must vanish, but there are clearly values of $\theta$ for which they are non-zero.