On the Definition of Taylor Polynomials One definition of Taylor Polynomials proceeds, as follows.

Definition:
Let $a \in \mathbb{R}$. Let $f$ be a function continuous at $a$. Let $n \in \mathbb{N}$. The $n$th Taylor Polynomial for $f$ at $a$ is the polynomial, $P_n$ of the smallest possible degree, which is an approximation of $f$ near $a$ of order $n$. That is,
$$\lim _{x \rightarrow a} \frac{f(x)-P_n(x)}{(x-a)^n}=0$$

I wish to ask why it was necessary to state of the continuity of $f$ at $a$.
Edit: Most namely, we know this limit to be satisfied when the first $n$ derivatives of $f$ and of $P_n$ agree. Yet, if we merely presume of $f$'s continuity at a point, how may guarantee that $f$ is $n$-times differentiable? Is it not more reasonable to presume of $f$'s being $C^n$ (by which I mean that all $n$ derivatives exist and are continuous)?
 A: For one, if $f$ is continuous at $a,$ then $\lim_{x \to a} f(x) = f(a).$ But also, the statement $$\lim_{x \to a} \frac{f(x) - P_n(x)}{(x - a)^n} = 0$$ says that the difference function $e(x) = f(x) - P_n(x)$ tends to $0$ more quickly than the polynomial $(x - a)^n$ as $x$ approaches $a.$ Observe that the function $e(x)$ gives the error in approximating $f(x)$ via the polynomial $P_n(x),$ hence this limit gives a criterion for the approximation to be "good enough" in some rigorous sense.
A: Let's assume that we don't know if $f$ is continuous at $a$. We only assume that $\lim\limits_{x\to a}\frac{f(x)-P_n(x)}{(x-a)^n}=0$. Then for an arbitrary $\frac{\epsilon}{2}>0$ we find a $\delta'>0$ (which depends on $a$) such that for all $x\in\mathbb{R}$ with $|x-a|<\delta'$ we have $\Bigl|\frac{f(x)-P_n(x)}{(x-a)^n}\Bigl|<\frac{\epsilon}{2}$. Let's define $\delta:=\min\{\delta',1\}$. Then it holds that:
$$\Bigl|\frac{f(x)-P_n(x)}{(x-a)}\Bigl|<\Bigl|\frac{f(x)-P_n(x)}{(x-a)^n}\Bigl|<\frac{\epsilon}{2}.$$
As polynomials are $n$-times differentiable we find a $\delta''>0$ for the above given $\frac{\epsilon}{2}$ such that for all $x\in\mathbb{R}$ with $|x-a|<\delta''$, we can conclude
$$\Bigl|\frac{P_n(x)-P_n(a)}{(x-a)}\Bigl|<\frac{\epsilon}{2}.$$
If we the set $|x-a|<\min\{\delta,\delta''\}$ it yields:
$$
\Bigl|\frac{f(x)-P_n(a)}{(x-a)}\Bigl|\leq \Bigl|\frac{f(x)-P_n(x)}{(x-a)}\Bigl|+\Bigl|\frac{P_n(x)-P_n(a)}{(x-a)}\Bigl|<\epsilon.
$$
If we then assume that approximation means $P_n(a)=f(a)$ then $f$ would be differentiable at $a$ with $f'(a)=0$. Hence, $f$ is always continuous at $a$ - you don't have to assume it. Without any further assumptions one cannot say if $f$ has higher order derivatives at $a$.
