Is f totally differentiable at (0,0)?

[springo]

Homework Statement

For any (x,y) other than (0,0).
f(x,y)=\tan(x·y)\sin\left(\frac{1}{x^2+y^2}\right)
For (x,y) = (0,0)
f(x,y) = 0
Is f totally differentiable?

Homework Equations

The Attempt at a Solution

If the function is not continuous, it can't be differentiable.
\lim_{(x,y)\rightarrow(0,0)}\tan(x·y)\sin\left(\frac{1}{x^2+y^2}\right)=\lim_{(x,y)\rightarrow(0,0)}\frac{x·y}{x^2+y^2}=\lim_{r\rightarrow0}\cos(\theta)·sin(\theta)
So the limit doesn't exist.

However I looked at the solution and it is differentiable. Is it possible that the solution is wrong or did I make a mistake?

Thanks for your help.

Edit: I don't why it doesn't TeX (at least on my browser) so:
f(x,y) = tan(xy)*sin(1/[x^2+y^2])
And for my attempt at a solution:
lim(x,y->0,0) f(x,y) = lim (x,y->0,0) xy/(x^2+y^2) = lim (r -> 0) cos T*sin T

 

[tiny-tim]

Hi springo!

(for some reason the LateX doesn't seem to be working properly today … it's not just you! )

(and try using the X² tag just above the Reply box )

Is f totally differentiable?
…
If the function is not continuous, it can't be differentiable.
…
So the limit doesn't exist.
…
f(x,y) = tan(xy)*sin(1/[x^2+y^2])
And for my attempt at a solution:
lim(x,y->0,0) f(x,y) = lim (x,y->0,0) xy/(x^2+y^2) = lim (r -> 0) cos T*sin T

erm … sin(1/r) is not approximately 1/r, it's ≤ 1,

so |f(x,y)| ≤ tan(xy) -> 0

 

[springo]

Hi tiny-tim! Thanks for your answer.
I'm not sure I understand what you meant with your answer.
I was trying to find the limit by using polar coordinates:
x = r·cos(t)
y = r·sin(t)

So lim_(x,y)->(0,0) tan(x·y)·sin(1/[x²+y²])
~ lim_(x,y)->(0,0) x·y/(x²+y²)
~ lim_r->0 r²·cos(t)·sin(t)/r²
~ cos(t)·sin(t)
And so the limit doesn't exist.

 

[tiny-tim]

So lim_(x,y)->(0,0) tan(x·y)·sin(1/[x²+y²])
~ lim_(x,y)->(0,0) x·y/(x²+y²)

Noooo … as r -> 0, sin(r) -> r but sin(1/r) does not -> 1/r, it stays ≤ 1 …

try drawing it!

 

[springo]

OK, I understood! How could I make that mistake...
So how can I do this?

Edit:
I was thinking:
lim_h->0 [f(x,y) - f(0,0) - f_x(0,0)(x-0) - f_y(0,0)(y-0)]/√(x²+y²)
Since f_x(0,0) = f_y(0,0) = 0 (by doing the math... lol) and f(0,0) = 0... it's now:
lim_h->0 f(x,y)/√(x²+y^y) ~ x·y·sin(1/x²+y²)/√(x²+y²)
And by turning into polar we have r²/r = r (the other stuff is bounded so...).
So the limit is 0, so it's differentiable.
Is that right?

 

[tiny-tim]

Hi springo!

Since f_x(0,0) = f_y(0,0) = 0 (by doing the math... lol) …

a bit complicated … and sometimes f_x(0,0) and f_y(0,0) exist when, along some non-axis curve, the derivative doesn't exist.

Just use the definition …

what is lim [f(x,y) - f(0,0)]/r ?