Why not include as a requirement that all functions must be continuous to be differentiable? 
Theorem: Suppose that  $f : A \to \mathbb{R}$ where $A \subseteq \mathbb{R}$. If $f$ is differentiable at $x \in A$, then $f$ is continuous at $x$.

This theorem is equivalent (by the contrapositive) to the result that if $f$ is not continuous at $x \in A$ then $f$ is not differentiable at $x$. 
Why then do authors in almost every analysis book, not take continuity of $f$ as a requirement in the definition of the derivative of $f$ when we (seemingly) end up with equivalent results?
For example I don't see why this wouldn't be a good definition of the derivative of a function

Definition: Let $A \subseteq \mathbb{R}$ and let $f : A \to \mathbb{R}$ be a function continuous at $a$. Let $a \in \operatorname{Int}(A)$. We define the derivative of $f$ at $a$ to be $$f'(a) = \lim_{t \to 0}\frac{f(a+t)-f(a)}{t}$$
  provided the limit exists.

I know this is probably a pedagogical issue, but why not take this instead as the definition of the derivative of a function? 
 A: Beyond pedagogy, there is a reason to avoid this restriction. It turns out that it's not strictly required and there are definitions of the derivative that can be computed on non-continuous functions. Instead of taking the definition of the derivative to be the difference equation you provided, if you instead use the fundamental theorem of calculus, you can get an alternate definition of the derivative. Specifically, if $F(x) = \int_{-\infty}^{x}f(x)dx$, then $f(x)$ is the derivative of $F(x)$, and does not require strict continuity. This definition amounts to "The stuff that when integrated yields big F".
There are alternative version of the derivative that handle more degenerate cases: e.g. http://mathworld.wolfram.com/Radon-NikodymDerivative.html
A: Because then you would need to check continuity for no good reason every time you want to check for differentiability. Besides, it gives the wrong impression of being necessary to include.
A: If we started to include consequences of definitions, in the definitions themselves, there's no telling where to stop.
A: The "obvious" (in Cauchy's or Riemann's view) properties of differentiation or continuity lead to the current definitions. The epsilon-delta definition of continuity seems reasonable; it's a bit difficult to find something else. This definition equates to "can draw without lifting the pencil."
The idea of differentiable has to do with slopes. One checks (f(x)-f(x-h))/h for existence and other properties. Then the result follows.
A: One possible reason is that the relationship between differentiability and continuity is more subtle in multivariable calculus.
Consider these definitions:

Let $A \subseteq \mathbb{R}^2$ and $f : A \to \mathbb{R}$ be a function. Let $a \in \operatorname{Int} A$. We define the directional derivative of $f$ at $a$ along the unit vector $v \in \mathbb{R}^2$ like this:
$$\partial_vf(a) = \lim_{h \to 0} \frac{f(a+hv) - f(a)}{h}$$
Futhermore, we say that $f$ is differentiable at $a$ if there exists a linear map $L :\mathbb{R}^2 \to \mathbb{R}$ such that 
$$\lim_{h\to 0} \frac{\left|f(a+h) - f(a) - Lh\right|}{\|h\|} = 0$$
  Note that $h \in \mathbb{R}^2$ here.

It can be shown that if $f$ is differentiable at $a$ then $L$ is unique and the directional derivatives exists along any unit vector $v \in \mathbb{R}^2$, being equal to $\partial_vf(a) = Lv$. Also, it implies that $f$ is continuous at $a$.
However, the converse is not true: if $f$ poseses directional derivatives along all unit vectors, $f$ does not even need to be continuous at $a$ (let alone differentiable):
Consider $f : \mathbb{R}^2 \to \mathbb{R}$ given by
$$f(x,y) = \begin{cases}
1,  & \text{if $0 < y < x^2$} \\
0, & \text{otherwise}
\end{cases}$$
All directional derivatives at $(0,0)$ exist and are equal to $0$, but the function fails to be continuous at $(0,0)$.
A: Another reason is that, later on when you want to generalize, you might have to explicitly remove that requirement... why risk that when you never needed to have it in the first place?
(Example: the derivative of the unit step is the Dirac delta, neither of which is continuous at zero.)
A: Because that suggests that there might be functions which are discontinuous at  $a$ for which it is still true that the limit$$\lim_{t\to0}\frac{f(a+t)-f(a)}t$$exists. Besides, why add a condition that always holds?
A: Definitions tend to be minimalistic, in the sense that they don't include unnecessary/redundant information that can be derived as a consequence.
Same reason why, for example, an equilateral triangle is defined as having all sides equal, rather than having all sides and all angles equal.
A: With the standard definitions, it is an important theorem that differentiable implies continuous. As a theorem, it has real content: if a function can be locally approximated (in a way that can be made precise) at a point by a linear function then it is continuous at that point. With your proposed definition, differentiable implies continuous is a boring tautology. Of course, you could still formulate the non-boring version of the theorem so that it uses the revised terminology, but the result would be somewhat cumbersome. The standard definition allows you to elegantly phrase an important theorem. Your proposed revision obscures that elegance for no real gain.
A: I imagine that a large part of this is just tradition: that's how someone in the past wrote the definition, and so people continue to write it that way.
I imagine the rest boils down to the issue of teaching introductory calculus to a class of mathematically unsophisticated students. First teaching them that a continuity condition is part of the definition of "derivative" and later teaching them that the limit existing is sufficient on its own is going to lead to students getting confused and frustrated. Two specific negative things I imagine this will introduce are:


*

*It's already difficult to get students to pay attention to all of the hypotheses of a theorem or definition. Giving them an example of "here's a hypothesis... whoops it doesn't matter" in the very basics is likely to just reinforce that difficulty

*Some students will latch tightly around "the continuity condition is part of the definition", and will continue to do a lot of wasted work checking the condition, even after you teach them the limit existing is sufficient. They will also be suspicious of work by others (which includes the teacher and the textbook!) that doesn't check this condition every time. Or they will become disillusioned and turn into another example of the previous bullet point.

A: To add another perspective: this rewording of the definition has a kind of circularity to it. We know that differentiable functions must be continuous, so we define the derivative to only be in terms of continuous functions. But then, the fact that differentiable functions are continuous is by definition, while it is being used to justify that very definition. The only reason we were able to start this cycle of reasoning is because we know that, using the standard definition without the continuity requirement, we can prove that differentiable functions are continuous. Thus, we should stick with that definition to avoid such a situation.
