Question about a property that the derivative satisfies I understand the derivative as a limiting process, and I understand that  
$$f^{\prime}(a)=\lim_{h \to0}\frac{f(a+h)-f(a)}{h} \tag{1}$$
But I'm slightly confused about this property that the derivative satisfies, which I read is equivalent: 
$$\lim_{h \to0}\frac{f(a+h)-f(a)-f^{\prime}(a)h}{h}=0. \tag{2}$$
"which has the intuitive interpretation that the tangent line to $f$ at $a$ gives the best linear approximation:  
$$f(a+h) \approx f(a)+f^{\prime}(a)h \tag{3}$$
to $f$ near $a$ for small $h$." (Wikipedia).
I'm think with pictures. I can picture secant lines approaching a tangent line, and thus I understand the limit definition of the derivative in $\textbf{(1)}$.  
But I'm just confused about this particular property. I think I'm confused about the notation too and why $f^{\prime}(a)h$ is being subtracted from $f(a)$ in $\textbf{(2)}$ and added to $f(a)$ in $\textbf{(3)}$.  
Can you please explain and elaborate on this derivative property? Thank you. 
 A: In pictures, what this is trying to say is that if you are on the graph of a function $f$ at a point $a$, then you can approximate the value of the function at a small distance $h$ away from $a$, ($f(a+h)$) by using the derivative, $f^\prime(a)$.  $f(a) + f^{\prime}(a)h$ is the equation of the tangent line using $h$ as your independent variable.  
This type of approximation becomes common when you start talking about numerical solutions to differential equations.  
A: It's just a subtraction of limits!
\begin{align*}
f^{\prime}(a)&=\lim_{h \to0}\frac{f(a+h)-f(a)}{h} \\
f^{\prime}(a)&=\lim_{h \to0}\frac{hf'(a)}{h} \\
\end{align*}
Therefore, subtracting,
$$
\lim_{h \to 0} \frac{f(a + h) - f(a) - hf'(a)}{h} = 0.
$$
A: Concerning your question "why $f′(a)h$ is being subtracted from $f(a)$ in (2) and added to f(a) in (3)". In (2) you have that $$\frac{f(a+h)-f(a)-f'(a)h}{h}=\frac{f(a+h)-\left(f(a)+f'(a)h\right)}{h}$$
so actually it is both times the same (it is added).
A: Just for better explanation purposes...you're right in trying to visualize the meaning of an algebraic expression: in some cases a geometric approach gives you better results, so take a look at this picture:

We denote $A$ as the $y$-value of $f(a)$, while $B$ represents the value of $f(a+h)$, and the black lines are all parallel or perpendicular to the main axis $x,y$. We want to give a geometrical meaning of $$f(a+h) \approx f(a)+f^{\prime}(a)h $$ By understanding that equation you'll ABSOLUTELY find it easy to understand $\lim_{h \to0}\frac{f(a+h)-f(a)-f^{\prime}(a)h}{h}=0$, so let's work on this picture...
Consider the right-angled triangle $\widehat{ACD}$: we have $\overline{CD} = \overline{AC}\tan \hat{CAD} = h\cdot f'(a)$, in fact $f'(a)$ is the angular coefficient of the tangent at the function in the point $a$. So we have $$\overline{CD} = f'(a)h$$ and adding $f(a)$ in both sides of the equation we get $$\overline{CD} + f(a)= f(a)+f'(a)h$$ Note that if we consider a smaller $h$ we get a smaller $\overline{CD}$, and so $D \to B$. As a conclusion we can state $$\overline{CD} + f(a) = f(a)+f'(a)h \approx f(a)+ \overline{CB} = f(a+h)$$ for little $h$. In other words 
$$lim_{h \to 0} f(a+h)= f(a)+f'(a)h$$
