= Solution
<Taylor's theorem> expands about $a$ in powers of $h=x-a$. With the usual differentiability assumptions, its finite form with <Lagrange remainder> is
$$
f(x)=\sum_{k=0}^n\frac{f^{(k)}(a)}{k!}(x-a)^k+\frac{f^{(n+1)}(\xi)}{(n+1)!}(x-a)^{n+1},
$$
where $\xi$ lies between $a$ and $x$. If the remainder tends to zero as $n\to\infty$, this yields the <Taylor series> $f(x)=\sum_{k\geq0}f^{(k)}(a)(x-a)^k/k!$. Smoothness alone does not guarantee that infinite-series equality.
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