I've recently had to prove the following claim:
Let $ f: \mathbb{R} \rightarrow \mathbb{R} $ such that $\forall x_0 \in \mathbb{R} \space$ $ f^{(n)}(x_0)$ exists for each $ n \in \mathbb{N} $.
Suppose that $ \forall b \in (0, \infty) $ there exists some $ M \in \mathbb{R} $ s.t $\forall n \in \mathbb{N}$ and $\forall x \in [0, b] : | f^{(n)}(x) | \le M $.
Prove that $ \forall b \in (0, \infty) \space : \space \lim_{n \rightarrow \infty }P_{n}(b) = f(b) $ where $P_n(x) = P_{n,f,0}(x).$
My question is about the assumption here. According to my intuitive understanding, as n "gets bigger" as Taylor polynomial "behaves" more and more like $f$. I guess one's could come up with a counter example, but why my intuition isnt correct here? We do we have to assume the function's $n'th$ derivative values are always bounded? Feels like I misunderstand the concept.
Thanks
It is necessarily true that, as $n$ growths, the Taylor polynomials of a function $f$ at a point $a$ of its domain behave more and more as $f$. The standard counterexample is the function $f:\mathbb{R}\longrightarrow\mathbb{R}$ defined by $$ f(x)=\begin{cases} e^{-1/x^2}&\text{ if }x\neq0\\ 0&\text{ if }x=0. \end{cases} $$ Then every Taylor polynomial of $f$ at $0$ is the null polynomial.
However, for most $C^\infty$ functions $f$ that you will work with the Taylor polynomials of a function $f$ at a point $a$ of its domain will indeed behave more and more as $f$, but only near $a$. If, say you take $f(x)=\frac1{1+x^2}$, then the Taylor polynomials of $f$ at $0$ do get get closer and closer to $f$, but only in $(-1,1)$, not on the whole real line.
So, without some extra condition it will not be true that the Taylor polynomials of a function $f$ at a point $a$ of its domain behave more and more as $f$ on the whole real line. The problem that you stated gives one such condition.