Question

Let f(x, y) ={ y³ /x² + y² if (x, y) ≠(0, 0) and 0 otherwise

Show that f is continuous but not differentiable at (0, 0).

Answer 1

To show that the function

$$f(x,y)=\begin{cases} \frac{y^3}{x^2+y^2}~~~~(x,y) \neq (0,0)\\ 0~~~~~~~~~~(x,y)=(0,0)\end{cases}$$

is continuous at $(0,0)$ , we must show that $\lim_{9x,y)\to (0,0) }f(x,y)=f(0,0)$

By definition, $f(0,0)=0$.

Let $y=mx$ . Thus,

$\lim_{(x,y)\to (0,0)} f(x,y)=\lim_{x \to 0}f(x,mx)=\lim_{x\to 0} \frac{(mx)^3}{x^2+(mx)^2}=\lim_{x \to 0} \frac{m^3x}{1+m^2}=0$

Thus the limit is 0 which is equal to $f(0,0)$

Hence, $f(x,y)$ is continuous at $(0,0)$

Now, to show that it is not differentiable at (0,0)

Differentiating with respect to x yields:

$$f_x(x,y)=\frac{-2xy^3}{(x^2+y^2)^2}$$

Now, $f_x(0,0)=0$ , but

Consider the limits along the straight line $y=mx$

$\lim_{(x,y)\to (0,0)}f_x(x,y)=\lim_{x\to 0} \frac{-2x(mx)^3}{(x^2+(mx)^2)^2}= \lim_{x\to 0}\frac{-2m}{(1+m^2)^2}=\frac{-2m}{(1+m^2)^2}$

Since the limit along a straight line depends on the slope of the line. Therefore, the two-variable limit does not exist at (0, 0). Since the limit does not exist, this shows that the partial derivative of the function with respect to x exist but is not continuous.This shows that the function is not differentiable.