Tangent plane to the graph of $f(x,y)=|xy|$ at $(0,0)$

Question

Given the function $$f(x,y)=|xy|$$ how to know if the graph of f has a tangent plane at $(0,0)$?

I know that $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ don't exist at $(0,0)$, is this enough to say that the tangent plane doesn't exist at that point?

My intuition says that maybe a tangent plane could exist since both axis are in the graph.

Answer 1

By linearization we have that the tangent plane at $p = (a,b,f(a,b))$ is given by;

$$z = f(a,b) + (x-a)f_x(p) + (y-b) f_y(p)$$

i.e you have $\langle f_x(p), f_y(p), -1 \rangle \cdot \langle x-a,y-b,z-f(a,b)\rangle = 0$ i.e $\langle f_x(p),f_y(p), -1 \rangle$ is the normal vector for the plane at $p$.

$$\frac{\partial f}{\partial x}(0,0) = \lim_{t \to 0} \frac{f(t,0) - f(0,0)}{t} = 0$$

$$\frac{\partial f}{\partial y}(0,0) = \lim_{t \to 0} \frac{f(0,t) - f(0,0)}{t} = 0$$

Answer 2

First, $$\frac {\partial f}{\partial x}\bigg|_{(x,y)}=\lim_{h\to\infty} \frac{f(x+h,y)-f(x,y)}{h}=\lim_{h\to\infty} \frac {| (x+h)(y)|- |xy|}{h}=\lim_{h\to\infty} |y| \frac{|x+h|-|x|}{h}$$ It follows that $$\frac {\partial f}{\partial x}\bigg|_{(0,0)}=\lim_{h\to\infty} |0| \frac{|0+h|-|0|}{h}=\lim_{h\to\infty} |0| \frac{|h|}{h}=0$$

Also, $$\frac {\partial f}{\partial y}\bigg|_{(x,y)}=\lim_{h\to\infty} \frac{f(x,y+h)-f(x,y)}{h}=\lim_{h\to\infty} \frac {| (x)(y+h)|- |xy|}{h}=\lim_{h\to\infty} |x| \frac{|y+h|-|y|}{h}$$

so $$\frac {\partial f}{\partial y}\bigg|_{(0,0)}=\lim_{h\to\infty} |0| \frac{|0+h|-|0|}{h}=\lim_{h\to\infty} |0| \frac{|h|}{h}=0$$

This means that both $$\frac {\partial f}{\partial x}\bigg|_{(0,0)}$$ and $$\frac {\partial f}{\partial y}\bigg|_{(0,0)}$$ exist.

It remains to show that $$\lim_{(x,y)\to(0,0)}\frac{f(x,y)-f(0,0)-\frac {\partial f}{\partial x}\bigg|_{(0,0)}(x-0)-\frac {\partial f}{\partial y}\bigg|_{(0,0)}(y-0)}{\|(x,y)\|}=0$$

But $$\lim_{(x,y)\to(0,0)}\frac{f(x,y)-f(0,0)-\frac {\partial f}{\partial x}\bigg|_{(0,0)}(x-0)-\frac {\partial f}{\partial y}\bigg|_{(0,0)}(y-0)}{\|(x,y)\|}$$ $$=\lim_{(x,y)\to(0,0)}\frac{f(x,y)}{\|(x,y)\|}$$ $$=\lim_{(x,y)\to(0,0)}\frac{|xy|}{\|(x,y)\|}$$ $$=\lim_{(x,y)\to(0,0)}\frac{|xy|}{\sqrt{x^2+y^2}}=0$$.

So with all this we can conclude that that $f$ has a tangent plane at $(0,0)$