Definition of differentiability at one point My Multivariable Calculus notes state that, in order for a function of the type $\mathbb{R}^2\rightarrow \mathbb{R}$ to be differentiable at a given point, the limit $$\lim _{h,\:k\to 0}\left(\frac{E\left(h,\:k\right)\:}{\:\sqrt{h^2+k^2}}\right)=0$$ Where $$f\left(a+h,\:b+k\right)-f\left(a,\:b\right)=h\cdot \frac{\partial }{\partial x}\left(a,\:b\right)+k\cdot \frac{\partial }{\partial y}\left(a,\:b\right)+E\left(h,\:k\right)$$ But everyone just expect us to memorize this and reproduce it on the exam. What does this really mean? Why does the limit must approach $0$?
 A: The essence of differentiation is in linear approximation (I use the term "linear" here as in polynomials, not in the sense of linear maps from linear algebra. In linear algebra terms, I'm talking about affine maps). In order for a function $f : \Bbb{R}^n \to \Bbb{R}^m$ to be differentiable at a point $a$, we require there to be some kind of linear function $L : \Bbb{R}^n \to \Bbb{R}^m$ that approximates $f$ at the point $a$, for a particular definition of "approximates".
Think about the $f : \Bbb{R} \to \Bbb{R}$ case. We define
$$f'(x_0) = \lim_{h \to 0} \frac{f(x_0 + h) - f(x)}{h},$$
giving us a number in $\Bbb{R}$, when defined. We then use this number to generate a linearisation of $f$ at $(x_0, f(x_0))$:
$$L(x) = f'(x_0)(x - x_0) + f(x_0).$$
What I'd like you to do is flip this perspective, and think about $L(x)$ as the derivative, instead of the number $f'(x_0)$. Let's suppose we have function $T(x) = mx + b$ that approximates $f$ in the following sense:
$$\frac{|f(x_0 + h) - T(x_0 + h)|}{|h|} \to 0$$
as $h \to 0$. Then we can actually deduce that $f'(x_0)$ exists, and that $T = L$. That is, if a linear approximation exists in the above sense, then it proves differentiability in the traditional sense. If you note that $T = L$ satisfies the above limit, then you can see that linear approximations existing in this sense is equivalent to differentiability at $x_0$.
Note the similarity of the above limit to the limit
$$\frac{E(h, k)}{\|(h, k)\|} \to 0$$
as $(h, k) \to (0, 0)$. In the $\Bbb{R}^2 \to \Bbb{R}$ case, we simply want a linear function $T(x, y) = px + qy + r$ that satisfies
$$\frac{|f(a + h, b + k) - T(a + h, b + k)|}{\|(h, k)\|} \to 0.$$
If such a $T$ exists, we can deduce that $p$ is the $x$-partial derivative at $(x, y)$, $q$ is the $y$-partial derivative, and $r = f(a, b)$. So, this comes to be the expression (up to a missing absolute value) of which you need the limit.
So, in short, derivatives are linearisations, rather than numbers. The limit definition you've been given guarantees a linear approximation to the function at $(a, b)$, where the (one and only possibility for the) linear function is given by
$$L(x, y) = \left(\frac{\partial f}{\partial x}(a, b)\right) \cdot h + \left(\frac{\partial f}{\partial y}(a, b)\right) \cdot k + f(a, b).$$
A: For a fixed point $(a,b)$ the function
$$
    L(a+h,b+k) = f(a,b) + h \frac{\partial f}{\partial x}(a,b) + k \frac{\partial f}{\partial y}(a,b)
$$
is a (affine) linear function that satisfies $L(a,b) = f(a,b)$, $\frac{\partial L}{\partial x}(a,b) = \frac{\partial f}{\partial x}(a,b)$, and $\frac{\partial L}{\partial y}(a,b) = \frac{\partial f}{\partial y}(a,b)$.  So $L$ is essentially the best possible linear approximation to $f$ near $(a,b)$.
But how good an approximation is it?  That depends how close $f$ is to $L$ in a neighborhood of $(a,b)$.  We want the difference $f-L = E$ to go to zero as $(h,k) \to (0,0)$.  But not just go to zero, go to zero faster than $h$ and $k$ go to zero.  Or equivalently, faster than the length $\sqrt{h^2+k^2}$ goes to zero. That is why we take the quotient and require it to have a limit of zero.
Heuristically, $f$ is differentiable at $(a,b)$ if the graph of $L$ matches our intuitive notions of “tangent plane to the graph of $f$” at $(a,b)$.  
