How to show that two definitions of differentiability in $\mathbb{R}^2$ are equivalent

In my Advance Analysis course, the instructer gave the definiton of differentiability as $$\lim_{(\delta x, \delta y) \to (0,0)} \frac{ f(x_0 + \delta x, y_0 + \delta y) - f(x_0, y_0) - [f_x(x_0, y_0)\delta x + f_y (x_0, y_0)\delta y] }{\sqrt{\delta x^2 + \delta y^2}} = 0$$

But, for example, in Apostol, it is given as

$$\lim_{\vec v \to 0} f(c + \vec v) - f(c) = \nabla f(c) \cdot \vec v$$ at page 348, and I cannot prove show that these two definitions are equivalent by starting from Apostol's definition.

I mean if we directly take $\vec v = (\delta x, \delta y)$ (since apostol does not assume $\vec v$ is unit), we will have a missing $1/||(\delta x, \delta y)||$ factor in the expression, and I couldn't find where I'm missing, so any hint or help is appreciated.

• I think the Apostol definition is incorrect in the sense that the left limit is always zero if $f$ is continous. – Netivolu Dec 4 '17 at 10:15
• I agree with @Netivolu. A correct equivalent definition of differentiability which is somewhat similar could be: There exists a function, $\epsilon(\vec v)$ s.t. $$f(c + \vec v) - f(c) = \nabla f(c) \cdot \vec v + \epsilon(\vec v)$$ and $\epsilon(\vec v)$ is continuous at $c$ and $\epsilon(c) = 0$. – John Don Dec 4 '17 at 11:19