The normal derivative formula is given by $lim_{x->0} (f(x+h)-f(x))/h$ my question is the more general formula $lim_{h->a} (f(x+h)-f(x))/h$ discussed for a not equal to $0$.
for example $g(x)=lim_{h->1} (f(x+h)-f(x))/h = f(x)-f(x+1)$ could this be used to approximate larger distances from a known value of x using newtons approximation method, i.e. $\sqrt{84}=9^2 + g(9)*2$ and $f(x)=\sqrt{x}$
In order to match the derivative definition, your limits should be for $h$, not $x$. When $h \to a \neq 0$ then you are calculating the average rate of change of the function between $x$ and $x + a$.