Is the following function differentiable?

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SUMMARY

The function in question is analyzed for differentiability at the point (0,2). The gradient is calculated as (\nabla f)(0,2) = (0,1), indicating a potential differential of \lambda(h,k) = k. The limit to prove differentiability is set up as \displaystyle\lim_{(h,k) \to (0,0)} \frac{|f(0+h,2+k)-f(0,2)-\lambda(h,k)|}{\left\|{(h,k)}\right\|}=0. The discussion emphasizes the need for continuity at (0,2) and the evaluation of partial derivatives for x approaching zero, which complicates the analysis. The suggestion to use Taylor expansion is noted but not fully explored.

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I have:
http://img12.imageshack.us/img12/6121/capturerhf.png


Is the function differentiable in (0,2)? If so, find its Tangent Plane.

So far I have
We have (\nabla f)(0,2)=(f_x(0,2).f_y(0,2))=\ldots=(0,1) , so if f is differentiable at (0,2) the only possible differential is \lambda (h,k)=(\nabla f)(0,2)(h,k)^t=k .So I have to analyze \displaystyle\lim_{(h,k) \to (0,0)} \frac{|f(0+h,2+k)-f(0,2)-\lambda (h,k)|}{ \left\|{(h,k)}\right\|}=0 but I can't seem to solve it.

And I also don't know how to find a tangent plane
 
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I came to a conclusion that f is indeed differentiable.

Can someone please help me understand how to prove it? Someone suggested that i use the Taylor expansion but i don't know how to use it. So I'm hoping someone could show me :)
 
As a prerequisite, the function needs to be continuous at the point in question for a derivative to exist at that point. Is this function continuous at (x,y)=(0,2)?

Assuming it is continuous, this is a function of two variables, so you need to investigate both the partial with respect to x and with respect to y. For x not equal to zero you will get one pair of partial derivatives, for x equal to zero you will get another. Is the first set equal to the second set in the limit x→0?
 
when i try to take the partial derivatives i can't really take them to x->0 and i get stuck with the epsilon business.

The general steps for showing differentiability doesn't really apply here.
 

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