I am copying this from my book.
Example: For function $f(x)=e^x,x\in \mathbb{R}$ and $c=0$ is $f^{(n)}(0)=1$, so Taylor series around zero are $\sum_{n=0}^{\infty} \frac{x^n}{n!}$. For $\delta\gt0$ we have $(x\lt \delta) \Rightarrow (e^x\lt e^{\delta})$, thus $\forall x\in \langle -\delta,\delta \rangle$ is $|e^x|\lt e^{\delta}=C$. Now, for every $x\in \mathbb{R}$, there exists $\delta \gt 0$, such that $x\in \langle -\delta,\delta \rangle$, so series converge to function on whole $\mathbb{R}$.
I understand first sentence of this example, but i don't know why existence of $\delta \gt 0$ such that... leads to conclusion that series converge to function on whole $\mathbb{R}$
I can't read the text, but looking at the bottom of the previous page, I find a formula for the remainder term in the $n$th order Taylor polynomial. Given that all higher derivatives of $e^x$ are $e^x$, and the boundedness of $e^x$ on bounded intervals, it does indeed follow that the Taylor series converges to the function (i.e., that the remainder term goes to zero). The author is just leaving the details of that calculation to the reader.
Can you take it from there yourself now?