from Carathéodory Derivative definition to the derivative of $\sin(x)$ A function $f$ is Carathéodory differentiable at $a$ if there exists a function $\phi$ which is continuous at a such that
$$f(x)-f(a)=\phi(x)(x-a).$$ 
For $f(x) = x^n$, $\phi(x)  = x^{n-1} + ax^{n-2} + ... + a^{n-1}$. We can see that $f'(a) = \phi(a) = na^{n-1}$. We get the derivative of $x^n$ directly from this definition. 
For $\sin(x)$, can we do the same thing to get $\sin'(x) = \cos(x)$ without using limit?
My question is raised from the booklet Calculus for Mathematicians by D.J.Bernstein, in which he defined derivative this way before introducing the concept of limits.  
 A: Because of definition of $\sin(x)$, it is very unlikely that we will find a nice representation of $\phi$ by elementary functions. This is why it isn't as easy as for $x^n$:
First, we recall that $\sin(x)$ is analytically defined as infinite series, $\sum\limits_{i=0}^\infty \frac{(-1)^n}{(2n+1)!}x^{2n+1} $. Then, 
$$ \sin(x) - \sin(a) = \sum\limits_{i=0}^\infty \frac{(-1)^n}{(2n+1)!}(x^{2n+1}-a^{2n+1}) = (x-a) \cdot \sum\limits_{i=0}^\infty \frac{(-1)^n}{(2n+1)!}(x^{2n} + ax^{2n-1} + ... + a^{2n}) = (x-a) \cdot \phi(x).$$
Also, we have $\phi(a) = \sum\limits_{i=0}^\infty \frac{(-1)^n}{(2n+1)!}((2n+1)a^{2n}) = \sum\limits_{i=0}^\infty \frac{(-1)^n}{(2n)!}a^{2n} = \cos(a). $
Notice that, unlike $x^n$-example, here (even if we overlook convergence problems) $\phi$ doesn't have a nice representation by elementary functions, so that's why it isn't as easy to do the same thing as for $x^n$ and get a Caratheodory derivative. 
A: The Weierstrass-Caratheodory formulation asserts that $f\colon E\rightarrow\mathbb{R}$ is differentiable at a point $x_0\in E$ if and only if there exists a function $\phi$ continuous at $x_0$ such that 
$$f(x)=f(x_0)+\phi(x)(x-x_0)$$
and the derivative is given by $f^{\prime}(x_0)=\phi(x_0)$.
Take $f(x)=\sin(x)$, then suppose we have the following equation:
$$\sin(x)=\sin(x_0)+\phi(x)(x-x_0)$$
Rearranging gives us:
$$\phi(x)=\frac{\sin(x)-\sin(x_0)}{x-x_0}$$
As you can see, if you can show that this quotient has a limit as $x\rightarrow x_0$, then you are done. But this is exactly the same as finding the derivative of $\sin(x)$ from first principle.
I could go on and derive it, but it is an easy exercise to show that it has a limit by using double-angle formula :) 
A: I am more comfortable with this version, find a continous function q_x such that:
f(x+h)-f(x) = q_x(x+h)*h 

Then we have:
f'(x) = q_x(x)

If you try it with f(x)=sin(x), you get this here:
sin(x+h)-sin(x) = 2 cos(h/2 + x) sin(h/2)

                = cos(((x+h)+x)/2) sin(((x+h)-x)/2)/(((x+h)-x)/2) h

                = q_x(x+h) h

The only continuous function q_x(x+h) that does the job, is the
one where lim h->0 sin(((x+h)-x)/2)/(((x+h)-x)/2) = 1 is used.
So that we have:
f'(x) = q_x(x) = cos(x) 1 = cos(x)

According to Stephen Kuhn in his intro this lim connection seems to be 
inevitable. It might be evitable if f(x) is a simple polynomial, where we
might regress to some reminder theorem.
