I'm trying to understand differentiability for multivariable functions and am thoroughly confused by the introduction (and the direction of implications in a certain definition)
I'm given the following definition of differentiability, word-for-word:
A function f(x,y) is differentiable at (a,b) IF there is a linear function $$ L(x,y) = f(a,b) + c(x-a) + d(y-b) $$ such that $$ \lim_{(x,y)\rightarrow (a,b)} \frac{|f(x,y) - L(x,y)|}{||(x,y) - (a,b)||} = 0 $$
The presence of the 'if' suggests to me that I can flip the order of the clauses in the sentence, to say:
"IF this linear function exists (which is characterized as a good approximation of f near the given point), THEN f is differentiable."
But this isn't an if-and-only-if, so the converse isn't necessarily true. i.e. f being differentiable doesn't mean that this linear function exists.
Then most resources I see follow up with a theorem
If f is differentiable at (a,b), then the linear function $$ L(x,y) = f(a,b) + c(x-a) + d(y-b) $$ is identical to the linear approximation, where c = f_x (a,b) and d=f_y (a,b).
The proof of this is where I get confused, because the first step in all the resources I've seen start with
Since f is differentiable at (a,b), we have $$ \lim_{(x,y)\rightarrow (a,b)} \frac{|f(x,y) - L(x,y)|}{||(x,y) - (a,b)||} = 0 $$
But I thought this was the hypothesis of the definition, not the conclusion! The function is differentiable if such a linearization exists. Rephrased, "If such a linearization exists, the function is differentiable." This proof is using the converse by saying that if it's differentiable, the linear function with error->0 exists.
Am I misinterpreting the implication in the definition? I've seen the exact same wording in multiple definitions online. "f is differentiable IF", not "if f is differentiable THEN". My head hurts. Please save me from my misery.