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## Homework Statement

For any (x,y) other than (0,0).

[tex]f(x,y)=\tan(x·y)\sin\left(\frac{1}{x^2+y^2}\right)[/tex]

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.

[tex]\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)[/tex]

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

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