Definition of Differentiablility/Continuity f:R^2->R

[clearweave]

Definition of Differentiablility/Continuity f:R^2-->R

Good afternoon,

On a test I was given the question:
Define f:R^2-->R by f(0,0)=0 and f(x,y)=x^3/(x^2+y^2)
(1) Do the partial derivatives at x and y exist at 0? Which I answered yes to, and showed that they equal 1 and 0, respectively.
(2) Is f differentiable at 0? Justify your answer.
(3) Is f continuous at 0? Justify your answer.

For part (2) I had trouble getting anywhere by the definition of a derivative with lim (h1,h2)-->(0,0) of [f((0,0)+(h1,h2))-f(0,0)]/norm(h1,h2), so I continued to (3) the continuity problem and since lim (x1,x2)-->(0,0) of f(x1,x2)=0/0 and is thus undefined, I stated that the definition of continuity lim (x1,x2)-->(a1,a2) of f(x1,x2) = f(a1,a2) thus cannot hold, so f is not continuous at (0,0). Since differentiability implies continuity, I used the contrapositive argument for (2) referencing (3), and I did not receive any credit for problems (2) and (3).

I considered the L1 norm, the Euclidean norm, and the sup norm for (2) to no avail, so I figured that it was possible the function is not differentiable at (0,0) since existence of partial derivatives does not imply complete differentiability... But apparently that was not the case here. Could anyone please help me solve (2) and (3) successfully? Thank you!

 

[micromass]

For (3), you calculated the limit to be 0/0, this is correct, but it is not enough. when you get 0/0 (or any other indeterminate form), then the limit might still exist, but you'll need to find another method to solve the limit.

To prove continuity, we often use the following inequalities:

$$|x|\leq \sqrt{x^2+y^2},~\frac{x}{x+1}<1,~\frac{x^2}{x^2+y^2}<1$$.

Now, we need to find an delta such that

$$\sqrt{x^2+y^2}<\delta~\Rightarrow~\left|\frac{x^3}{x^2+y^2}\right|<\epsilon$$

We make the following estimation

$$\left|\frac{x^3}{x^2+y^2}\right|\leq |x|\left|\frac{x^2}{x^2+y^2}\right|\leq |x|\leq \sqrt{x^2+y^2}<\delta$$

So choose delta=epsilon, and the definition is satisfied. Thus the function is continuous.

For differentiability, we need to check the definition, that is, we must find a linear function u such that

$$\lim_{h\rightarrow 0}{\frac{f(h)-f(0)-u(h)}{\|h\|}}$$

From your theory, you should know that there is only one possible choice for u(x,y), namely

$$u(x,y)=x\frac{\partial f}{\partial x}(0)+y\frac{\partial f}{\partial y}(0)$$=x[/tex]

So, we need to check whether the limit

$$\lim_{(h_1,h_2)\rightarrow (0,0)}\frac{f(h_1,h_2)-f(0,0)-(h_1,h_2)}{\sqrt{h_1+h_2}}=\lim_{(h_1,h_2)\rightarrow (0,0)}\frac{h_2^3}{\sqrt{(h_1+h_2)^3}}$$

However, you can check that this is not the case. Thus the function is not differentiable...