The way my professor defined Taylor polynomials is: the $n^{th}$ degree Taylor polynomial $p(x)$ of $f(x)$ is a polynomial that satisfies $\lim_{x\to 0}{f(x)-p(x) \over x^n} = 0$. This is actually the little-o notation $o(x^n)$, which means $(f(x)-p(x)) \ll x^n$ as $x$ approaches $0$. From this I have got the intuition that Taylor Polynomials work only for $|x| < 1$ because $x^n$ gets smaller as $n$ gets bigger only when $|x| < 1$. And the textbook seemed to agree with my intuition, because the textbook says “Taylor polynomial near the origin” (probably implying $|x| < 1$).
Since Taylor Series is basically Taylor polynomial with $n\to\infty$, I intuitively thought that the Taylor Series would also only converge to the function it represents in the interval $(-1, 1)$.
For example, in the case of $1\over1-x$, it is well known that the Taylor series only converges at $|x| < 1 $.
However, all of a sudden, the textbook says that the Taylor series of $\cos x$ converges for all real $x$. It confused me because previously I thought the Taylor series would only work for $|x|<1$. Now, I know that the Taylor Series is defined like this: $$ f(x) = Tf(x) \Leftrightarrow \lim_{n\to\infty}R_{n}f(x) = 0 $$
And I know how to get the maximum of Taylor Remainder for $\cos x$ using Taylor's Theorem, and I know that the limit of that Taylor Remainder is $0$ for all real $x$, which makes the Taylor Series of $cosx$ converge to $\cos x$, pointwise. However, I just can't get why my initial intuition is wrong (why taylor series converges for all $x$ for certain functions, like $\cos x$, also $\sin x$ and $e^x$, etc.)