There are lots of different ways this can be thought about, though probably the simplest and most common way is to use the inequality |R_{k}(x)|\leq \frac{M\cdot r^{k+1}}{(k+1)!}, where the (k+1)st derivative f^{(k+1)}(x) of f satisfies |f^{(k+1)}(x)|\leq M over the interval [c-r,c+r] (assuming sufficient differentiability/smoothness of f over the interval).
The function R_{k}(x) is the "remainder term" and is defined to be R_{k}(x)=f(x)-P_{k}(x), where P_{k}(x) is the kth degree Taylor polynomial of f centered at x=a: P_{k}(x)=f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f'''(a)}{3!}(x-a)^3+\cdots.
For example, if f(x)=e^(x), a=0, and k=4, we get P_{4}(x)=1+x+\frac{x^{2}}{2}+\frac{x^{3}}{6}+\frac{x^{4}}{24}. Moreover, if we consider this over the interval [-2,2] so that r=2, then the inequality above becomes |R_{4}(x)|\leq \frac{e^{2}\cdot 2^{5}}{5!}=\frac{4}{15}e^{2}\approx 1.97. So we can expect P_{4}(x) to approximate e^(x) to within this amount over the interval [-2,2] (actually, the approximation is even better than the error bound gives...the point is that the error bound gives a guarantee ).
