In a calculus class we were given the following definition of "differentiable function" (working with 2 variables):
Definition: Let $A \in \mathbb{R^2}$, and $f : A \to \mathbb{R}$. We say that $f$ is differentiable in $(x_0, y_0) \in A$ if the graph of $f$ admits a tanget plane at $(x_0, y_0, f(x_0, y_0))$.
Then the teacher gave us the following equivalent characterization:
Proposition: $f$ is differentiable in $(x_0, y_0)$ iff
1) $f$ admits partial derivatives in $(x_0, y_0)$
2) the following holds: $ \lim_{(x,y) \to (x_0, y_0)} \frac{f(x,y) - f(x_0,y_0) - A(x-x_0) - B(y-y_0)}{\sqrt{(x-x_0)^2 + (y-y_0)^2}} = 0 $ where $A = \frac{\partial f}{\partial x}(x_0,y_0)$ $B = \frac{\partial f}{\partial y}(x_0,y_0)$ i.e. the partial derivatives evaluated in $(x_0,y_0)$
Unluckily, I was not able to find any reference about this.
So here's my questions:
- I got it right? Are the two definitions equivalents?
- How to prove that the limit is zero iff the function admits a tangent plane?
Isn't (2) quite obvious? If a tangent plane at $P$ exists, its equation has to be $f(x,y) = f(P) - \frac{\partial f}{\partial x}(P)(x-x_P) - \frac{\partial f}{\partial y}(P)(y-y_P)$ follow that the numerator of the limit is zero. So what's the point of the denominator? Couldn't one use anything else for the denominator? Am I missing something? Is (2) noteworthy?
EDIT: As noted below, my "it's obvious from eq. of tangent plane" approach is, in fact, wrong.