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 #k#th 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 ).
