- #1

lep11

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## Homework Statement

Let ##f##: ##G\subset\mathbb{R}^2\rightarrow\mathbb{R}## be differentiable at ##(x_0,y_0)\in{G}## and ## \lim_{(x, y) \to (x_0, y_0)} \frac{f(x,y) -a -b(x-x_0) -c(y-y_0)}{\sqrt{(x-x_0)^{2} + (y-y_0)^{2}}} = 0.## The task is to prove that then ##a=f(x_0,y_0),b=f_x(x_0,y_0)## and ##c=f_y(x_0,y_0)##.

## Homework Equations

## The Attempt at a Solution

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Since ##f## is differentiable at ##(x_0,y_0)##, there exists a linear function ##L## such that

$$\lim_{(h_1, h_2) \to (0, 0)} \frac{f(x_0 + h_1, y_0 + h_2) - f(x_0, y_0) - L(h_1,

h_2)}{\sqrt{h_1^{2} + h_2^{2}}} = 0,$$where

$$

L(h_1, h_2) = \frac{\partial f}{\partial x}(x, y)\, h_1 + \frac{\partial f}{\partial y}(x, y)\, h_2.

$$Now we can substitute ##L(h_1,h_2)## in and get

$$\lim_{(h_1, h_2) \to (0, 0)} \frac{f(x_0 + h_1, y_0 + h_2) - f(x_0, y_0) - \frac{\partial f}{\partial x}(x, y)\, h_1 - \frac{\partial f}{\partial y}(x, y)h_2}{\sqrt{h_1^{2} + h_2^{2}}} = 0,$$

which is equivalent to $$ \lim_{(x, y) \to (x_0, y_0)} \frac{f(x,y) -f(x_0,y_0) -\frac{\partial f}{\partial x}(x, y)(x-x_0) -\frac{\partial f}{\partial y}(x, y)(y-y_0)}{\sqrt{(x-x_0)^{2} + (y-y_0)^{2}}} = 0,$$because ##(x-x_0)\rightarrow{0}## and ##(y-y_0)\rightarrow{0}## as ##(x,y)\rightarrow(x_0,y_0)##.

Therefore it must (?) hold that ##f(x,y) -a -b(x-x_0) -c(y-y_0)=f(x,y) -f(x_0,y_0) -\frac{\partial f}{\partial x}(x, y)(x-x_0) -\frac{\partial f}{\partial y}(x, y)(y-y_0)##

which implies (?) that ##a=f(x_0,y_0),b=f_x(x_0,y_0)## and ##c=f_y(x_0,y_0)##. I am thinking that maybe my proof is wrong and based too much on intuition rather than formal mathematical reasoning. I am asking for comments and tips on how to possibly modify or rewrite it in case it's not correct. There may be other approaches to this problem as well. I will appreciate your answers.

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