= Solution
One standard finite form of <Taylor's theorem> is the following. Suppose $f$ has continuous <derivatives> through order $m+1$ on an <interval> containing $a$ and $x$. Then, if $x\ne a$, there is some $\xi$ strictly between them such that
$$
\boxed{f(x)=\sum_{k=0}^m\frac{f^{(k)}(a)}{k!}(x-a)^k+\frac{f^{(m+1)}(\xi)}{(m+1)!}(x-a)^{m+1}.}
$$
The final term is the <Lagrange remainder>. At $x=a$ the identity holds directly with zero <remainder>. The finite expansion requires no assumption that an infinite <Taylor series> converges to $f$.
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