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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:

  1. I got it right? Are the two definitions equivalents?
  2. 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.

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2 Answers 2

up vote 6 down vote accepted

The definition is stated in terms of "the graph of $f$ admits a tangent plane". Did your teacher preceded these words by defining precisely what it means for a graph to admit a tangent plane? If not, then this "definition" was meant as an intuitive description of the concept, not as a definition in mathematical sense.

In the proposition, part 2) is actually the key, and part 1) could even be omitted. Namely, one can state the definition as follows:

A function $f$ is differentiable at $(x_0,y_0)$ iff there exists numbers $A$ and $B$ such that $$\lim_{(x,y) \to (x_0, y_0)} \frac{f(x,y) - f(x_0,y_0) - A(x-x_0) - B(y-y_0)}{\sqrt{(x-x_0)^2 + (y-y_0)^2}} = 0$$

Using this definition one can prove (and this is a reasonable exercise to do) that both partial derivatives exist and are equal to $A$ and $B$, respectively.

One can also give a geometric interpretation of the limit being zero. The equation $g(x,y)=f(x_0,y_0) + A(x-x_0) + B(y-y_0)$ is an equation of a plane. The difference $f(x,y)-g(x,y)$ measures how much the graph of the function deviates from this plane in the vertical direction. The limit being zero means that the vertical deviation at $(x,y)$ is small compared to the horizontal distance between $(x_0,y_0)$ and $(x,y)$: one pictures this as a surface and a plane meeting "at zero angle", that is, being tangent to each other. (But if at this point you'll ask me to define what being tangent means, I'll only point you back at the limit: the graphs of $f$ and $g$ are tangent if $$\lim_{(x,y) \to (x_0, y_0)} \frac{f(x,y) - g(x,y)}{\sqrt{(x-x_0)^2 + (y-y_0)^2}} = 0$$ The limit is the key; geometry is optional but helpful.)

In your post you wrote down the equation of the tangent plane incorrectly, and, which is a more serious mistake, denoted it by the same letter $f$ as the function itself. This is what led you to the erroneous conclusion that "numerator is zero". I avoid this confusion by using the letter $g$ for the function that represents the tangent plane: $z=g(x,y)$ is my notation for the tangent plane.

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Ok, now things seems to be a little more clear! I'll focus on the definition with the limit and the geometric interpretation of that. Thanks for your answer. –  Haile Jul 9 '12 at 12:32

Suppose that $G\subseteq \mathbb{R}^n$ is open, and $f:G\rightarrow \mathbb{R}^m$. Then $f$ is differentiable at $x$ if there is some linear transformation $A\in {\rm Hom}_{\mathbb{R}}(\mathbb{R}^n, \mathbb{R}^m)$ so that

$$f(x + h) - f(x) = Ah + o(\|h\|).$$

This abstracts to infinite-dimensional settings. In standard coördinates, the matrix of the linear transformation is the matrix of partials.

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1  
-1: not an appropriate explanation at this level. –  user31373 Jul 9 '12 at 1:05
2  
How did you decide that? In fact, most books I know (Apostol's "Calculus" II, (8.7), page 258, 2nd Ed., Rudin's "Principles of Mathematical Analysis", def. 9.11 page 212, 3rd Ed., etc.) define a differentiable function exactly in this way, i.e. by means of a linear map. This shouldn't be advanced stuff for a student dealing with differentiable functions of several variables. –  DonAntonio Jul 9 '12 at 2:16
    
I appreciate your answer, but I'm not able to understand it completely. The course was a lightweight introdution to 2-variables calculus for CompSci undegrads; and the teacher followed a quite relaxed approach mostly based on ideas from one-variable calculus. I probably should have written that in my question.. –  Haile Jul 9 '12 at 12:36

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