TITLE: What is the definition of differentiability?
QUESTION [6 upvotes]: Some places define it as:

If the Left hand derivative and the Right hand derivative at a point are equal then the function is said to be differentiable at that point.

Others define it based on the condition of the existence of a unique tangent at that point.
Which one of these is correct? Or are both of them wrong? 
According to the first definition, the curve need not be continuous at that point and can have a point discontinuity or a hole, like this:

Moreover it doesn't stop a curve, with a jump discontinuity but with same slope on both sides of it, from being differentiable.
And lastly, if a function is not defined at a point, then is the function discontinuous there too?

REPLY [4 votes]: A more general definition of differentiability is: 
$$\text{Function $f:\mathbb R\to \mathbb R$ is said to be differentiable if $\exists\,a\in \mathbb R$ such that $\lim_{h\to 0}\frac{f(x+h)-f(x)-ah}{\lvert h\rvert}=0\;$}$$
It can be shown that this definition is equivalent to the conventional definition that relies on the existence of $\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$ (in fact, $a$ is just $\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$ here!).

This definition can be extended naturally to multivariate function. We can define a multivariate function $g:\mathbb R^n\to\mathbb R$ as differential if:
$$\text{$\exists\,\mathbf A(\mathbf X)\in\mathbb R^n$ such that $\lim_{\|\mathbf H\|\to 0}\frac{g(\mathbf X+\mathbf H)-g(\mathbf X)-\mathbf A(\mathbf X)\cdot\mathbf H}{\|\mathbf H\|}=0$}$$
Likely, $\mathbf A(\mathbf X)$ can be proved to be exactly $(\nabla g)(\mathbf X)$, i.e., if $g$ is differentiable at $\mathbf X$, then the first order partial derivatives exist at $\mathbf X$ under above definition.