Is $f(a)-f(b) \approx f^{\prime}(b)(a-b)$ for small differences and if so, why? As in the question, I asked myself whether $f(a)-f(b) \approx f^{\prime}(b)(a-b)$ holds true for small differences between a and b and if so, why that is the case. Thanks for every hint.
 A: If $f(x)$ is continuous and differentiable, then, on a small enough scale it can be approximately a line.
In which case $f(x+\delta) \approx f(x) + \delta f'(x)$ for small $\delta$
We could also invoke the mean value theorem.
There exists a $c\in (a,b)$ such that $f'(c) = \frac {f(b) - f(a)}{b-a}$
A: If $f'$ is assumed to be bounded then $f(a)-f(b)=f'(t) (a-b)$ for some $t \in (a,b)$ so $|f'(t) (a-b)-f'(b) (a-b)|\leq 2M|a-b|$ where $M$ is a bound for $|f'|$. 
A: That is essentially Taylor's theorem:
$\begin{align*}
f(x)
   &= f(a) + f'(a) (x-a) + R_1(x)
\end{align*}$
Here $R_1(x)$ is the residue, which for functions with continuous second derivative tends to zero as $x$ tends to $a$ (thus the error is small for $x$ near $a$).
A: Put $a=b+h$ and define $\epsilon(h)=f’(b)-\dfrac{f(b+h)-f(b)}{h}$ then $\epsilon(h)$ goes to $0$ when $h$ goes to $0$. In particular $f(b+h)=f(b)+hf’(b)+\underbrace{h\epsilon(h)}_{\text{error term}}$ which is exactly what you stated.
A: You have to be careful to state precisely what you are assuming and what you want to be true.  One interpretation of the statement might be that if $f$ is a differentiable function on an interval $(c,d)$ (or maybe on the whole real line $\mathbb R$), for every $\epsilon > 0$ there is $\delta > 0$ such that if $c < a < b < d$ with
$0 < b-a < \epsilon$, $1-\epsilon < \dfrac{f(a)-f(b)}{f'(b)(b-a)} < 1+\epsilon$.
  But that's not true: for example you'll run into trouble if $f'(b)$ can be $0$.
