Taylor's polynomial uniqueness proof - why are these limits inferable?

Question

So I'm viewing a short proof on the uniqueness of Taylor polynomials.

Uniqueness of Taylor polynomial:
Let $f:]a,b,[ \rightarrow \mathbb{R}$ $n$ times continuously differentiable and $x_0 \in ]a,b,[$. If $p$ is $n$th degree polynomial function for which

$$f(x)-p(x)=o(|x-x_0|^n), \space \text{when} \space x \rightarrow x_0$$ then

$$p(x)=T_{n, x_0} f(x)$$

Proof starts like:

$$\frac{p(x)-T_{n, x_0}f(x)}{(x-x_0)^n} = \frac{p(x)-f(x)}{(x-x_0)^n}+\frac{f(x)-T_{n, x_0}f(x)}{(x-x_0)^n}$$

then

$$\frac{p(x)-f(x)}{(x-x_0)^n} \rightarrow 0 \space, when \space x-x_0 \rightarrow 0$$

$$\frac{f(x)-T_{n, x_0}f(x)}{(x-x_0)^n} \rightarrow 0 \space, when \space x-x_0 \rightarrow 0$$

Now my problem is, how can one actually say that the above approach $0$ if the denominator approaches $0$?

Answer 1

The correct result requires even less hypotheses.

Uniqueness of Taylor Polynomial: If $f$ is defined in a certain neighborhood of $a$ such that $f^{(n)}(a)$ exists and $P(x)$ is a polynomial of degree $n$ such that $$f(x) - P(x) = o((x - a)^{n})$$ as $x \to a$ then $$P(x) = f(a) + f'(a)(x - a) + f''(a)\frac{(x - a)^{2}}{2!} + \cdots + f^{(n)}(a)\frac{(x - a)^{n}}{n!}$$

To avoid any confusion let us remind ourselves that the notation $f(x) = o(g(x))$ as $x \to a$ is just another way of saying that $\dfrac{f(x)}{g(x)} \to 0$ as $x \to a$. This is reason why those limits in your question are $0$.

The idea behind the proof of "uniqueness of Taylor polynmials" is simple. If $R(x)$ is a polynomial of degree $n$ such that $$R(x) = o((x - a)^{n})$$ as $x \to a$ then $R(x) = 0$ identically. Why?? We use induction on $n$ the degree of polynomial $R(x)$. Thus let us assume that if $R(x)$ is a polynomial of degree $(n - 1)$ and $R(x) = o((x - a)^{n - 1})$ then $R(x) = 0$ identically. The case for $n = 0$ is clearly true. If $R(x) = k$ is a zero degree polynomial and $R(x) = o(1)$ then clearly $k = 0$.

Now let $R(x)$ be a polynomial of degree $n$ and let $R(x)$ be expressed in powers of $(x - a)$ as $$R(x) = r_{0} + r_{1}(x - a) + \cdots + r_{n}(x - a)^{n}$$ and $R(x) = o((x - a)^{n})$. This means that $\lim_{x \to a}R(x) = 0$ and hence $r_{0} = 0$ and $R(x) = (x - a)R_{1}(x)$ where $R_{1}(x)$ is a polynomial of degree $(n - 1)$. Now $R(x) = (x - a)R_{1}(x) = o((x - a)^{n})$ implies that $R_{1}(x) = o((x - a)^{n - 1})$ and hence by induction hypothesis $R_{1}(x)$ is identically $0$ and therefore $R(x) = (x - a)R_{1}(x)$ is also identically $0$.

Now from Taylor's theorem we know that the Taylor polynomial $$T(x) = f(a) + f'(a)(x - a) + f''(a)\frac{(x - a)^{2}}{2!} + \cdots + f^{(n)}(a)\frac{(x - a)^{n}}{n!}$$ is such that $f(x) - T(x) = o((x - a)^{n})$ and we also have $f(x) - P(x) = o((x - a)^{n})$ therefore via subtraction we see that $P(x) - T(x) = o((x - a)^{n})$ and hence by what we have proved above polynomial $P(x) - T(x)$ is the zero polynomial and hence $P(x) = T(x)$.

Answer 2

By Taylor's Theorem, we know that $f(x)-T_{n, x_0}=o(|x-x_0|^n)$ and we are assuming that $f(x)-p(x)=o(|x-x_0|^n)$ as $x \rightarrow x_0$. Now use the definition $o(|x-x_0|^n)$: $F=o(|x-x_0|^n)$ iff $$\lim_{x\to x_0}\frac{F(x)}{(x-x_0)^n}=0.$$

By the way, uniqueness can be proved without Taylor's Theorem. Assume that $p$ and $q$ are two polynomials of degree $\leq n$ such that $$f(x)-p(x)=o(|x-x_0|^n)\quad\mbox{and}\quad f(x)-q(x)=o(|x-x_0|^n).$$ Hence $r(x)=p(x)-q(x)$ is a polynomial of degree $\leq n$ and $r(x)=o(|x-x_0|^n)$. Then $r(x)=\sum_{k=0}^n a_k(x-x_0)^k$ and it is easy to see that $$0=\lim_{x\to x_0}\frac{r(x)}{(x-x_0)^n}=\lim_{x\to x_0}\frac{\sum_{k=0}^n a_k(x-x_0)^k}{(x-x_0)^n}$$ implies that $a_0=a_1=\cdots=a_n=0$, which means that the polynomials $p$ and $q$ are the same.