Assuming that " slope of the tangent at $x=a$ " and " $f'(a)$" are not identical by definition , how can their identity be shown algebraically? 
*

*When the concept of the " derivative number" of a function $f$ for some $x-$value , say $x=a$ is introduced - this number is simply $f'(a)$ -  the motivation is often that this number is identical to the slope of the tangent to the graph of $f$ at point $(a, f(a)$. 

*This is shown visually, by allowing a line passing through point $A =(a, f(a))$ and through some point $B$ on the curve to move gradually ( while $B$ also moves) until this line becomes identical to the tangent ( the slope of which we are supposed to look for). 

*My question is : how to show analytically that the number $\lim_{h\rightarrow0}\frac {f(a+h) - f(a)}{h}$ and the number $\frac{y_{D} - y_C} {x_{D} - x_C}$ (with $C$ and $D$ two arbitrary distinct points on the tangent line) are actually one and the same number? 

*To put it more briefly: assuming there is at least a conceptual difference between " derivative number" and " slope of the tangent" , how to show that the two expressions denote, in fact, one and the same object ( namely, the same number) ? 

*PS : here I use the expression " derivative number" that is common in french mathematics to denote the image of an x-value under the derivative function. 
 A: Here's one way to think of this.
Consider a function $y=f(x)$ defined on (say) an open interval $I \subset \mathbb R$, and consider $a \in I$ with value $f(a)$. 
Given a 1st degree polynomial $p(x) = cx+d$, let's say that $p(x)$ is a best linear approximation to $f(x)$ at $x=a$ if the following equation is true:
$$\lim_{h \to 0} \frac{f(a+h)-p(a+h)}{h} = 0
$$
What does this equation tell us? 
Intuitively, the "best linear approximation" is the idea behind the tangent line. First, the graphs of $y=f(x)$ and $y=p(x)$ pass through the same point $(a,f(a))$, i.e. $f(a)=p(a)$. Moreover, for values of $x$ near $a$ the difference between $f(a+h)$ and $p(a+h)$ is very small compared to $h$. So, if you were to "scale up" those graphs by a factor of $1/h$ and observe what happens --- as if peering through a more and more powerful microscope centered at the point $(a,f(a))$ --- then as $h$ gets closer and closer to $0$ the graph of $y=f(x)$ appears more and more like the graph of the straight line $y=p(x)$. 
To go a little bit deeper, one can also make formal deductions from this notion of "best linear approximation". 
First of all, it tells us that $f(a) = p(a)$, because both $f(x)$ and $p(x)$ are continuous at $x=a$ and therefore
\begin{align*}
f(a) - p(a) &= \lim_{h \to 0} (f(a+h)-p(a+h)) \\
 &= \lim_{h \to 0} \biggl(\frac{f(a+h)-p(a+h)}{h} \cdot h\biggr) \\
&= \lim_{h \to 0} \frac{f(a+h)-p(a+h)}{h} \cdot \lim_{h \to 0} h \\
&= 0 \cdot 0 = 0
\end{align*}
Second of all, it tells us that $f'(a)=p'(a)$, because
\begin{align*}
f'(a) - p'(a) &= \lim_{h \to 0} \frac{f(a+h)-f(a)}{h} - \lim_{h \to 0} \frac{p(a+h)-p(a)}{h} \\
 &= \lim_{h \to 0} \biggl( \frac{f(a+h)-f(a)}{h} - \frac{p(a+h)-p(a)}{h}  \biggr) \\
  &= \lim_{h \to 0} \frac{f(a+h)-p(a+h) - (f(a)-p(a))}{h} \\
  &=  \lim_{h \to 0} \frac{f(a+h)-p(a+h)}{h} \\
   &= 0
\end{align*}
Putting this together, we get $f(a)=p(a)=ca+d$, $f'(a)=p'(a)=c$. 
Solving the equations $f(a)=ca+d$ and $f'(a)=c$ for the quantities $c$ and $d$ and plugging in we get
$$p(x) = f(a) + f'(a) \cdot (x-a)
$$
which is exactly the function whose graph is the tangent line
$$y = f(a) + f'(a) \cdot (x-a)
$$
