We have $\displaystyle (\nabla f)(0,2)=(f_x(0,2).f_y(0,2))=\ldots=(0,1)$ , so if $\displaystyle f$ is differentiable at $\displaystyle (0,2)$ the only possible differential is $\displaystyle \lambda (h,k)=(\nabla f)(0,2)(h,k)^t=k$ . Now, analyze if $\displaystyle \displaystyle\lim_{(h,k) \to (0,0)} \frac{|f(0+h,2+k)-f(0,2)-\lambda (h,k)|}{ \left\|{(h,k)}\right\|}=0$ .
I came to that lim term myself and that's where I kind of get stuck with all the epsilons. I can't draw a concrete conclusion from there if it is indeed 0 or just a small epsilon bigger than 0. That's what making this exercise difficult.
I came to a conclusion that the function is indeed differentiable, but I have to show a strict rigorous proof.
How do I expand f to its Taylor Series?
And from there, how do I find the tangent plane?
That is a correct conclusion.
There are many ways to show a rigorous proof.but I have to show a strict rigorous proof.
You only need $\displaystyle \sin h(2+k)=h(2+k)-\frac{h^3(2+k)^3}{3!}+\ldots$ for finding in a comfortable way the limit in my answer #2.How do I expand f to its Taylor Series?
Use the well known formula $\displaystyle f_x(P_0)(x-x_0)+f_y(P_0)(y-y_0)+f_z(P_0)(z-z_0)=0$ where $\displaystyle P_0(x_0,y_0,z_0)$ belongs to the surface.And from there, how do I find the tangent plane?
Sorry for the follow up questions. Even with the Taylor expansion, how do I use it in the lim term you prescribed earlier?
And also, after i prove differentiability, how do I find the appropriate P0 to plug into the formula for the tangent plane?
I'm sorry for asking pathetic questions