Baby Rudin Theorem 8.2 This is Baby Rudin Theorem 8.2 proof:


In the last part of the proof, I don't understand why the following should be true for proof to work: $x>1-\delta$.
Also, what does it have to do with $-1<x<1$?
Any help is appreciated!
 A: Here, $|x|<1$.  So, $\Sigma_{n=0}^N |s_n-s||x|^n< \Sigma_{n=0}^N |s_n-s|$. This can be bounded by, say, $M$ because there are only a finite number of terms. So, choosing $\delta<\epsilon/2M$ gives us that $\color{blue}{(1-x)\Sigma_{n=0}^N |s_n-s| < (1-(1-\delta))\Sigma_{n=0}^N |s_n-s|} < \delta M <\epsilon/2$ which was what was desired. We have used the fact that $x>1-\delta$ in the blue inequality.
A: I think the following will explain what you wanted to know.
$
|f(x)-s|=|(1-x) ∑_{n=0}^\infty\;(s_n-s) x^n |$
$$≤|(1-x) ∑_{n=0}^N\;(s_n-s) x^n |+|(1-x) ∑_{n=N+1}^\infty\;(s_n-s) x^n |
$$
$$≤|(1-x) ∑_{n=0}^N\;(s_n-s) x^n |+ε/2\;|(1-x) ∑_{n=N+1}^\infty\; x^n| $$
$$≤|(1-x) ∑_{n=0}^N\;(s_n-s) x^n|+ε/2,\;\;\; ∀ε>0  \text{ and }  |x|<1.$$
If $x>1-\delta$, i.e., $1-x<\delta$, for some suitably so chosen $δ>0$ that $$δ\;|∑_{n=0}^N\;(s_n-s)| \le ε/2,$$
we obtain
$$|f(x)-s|\le\varepsilon/2+\varepsilon/2=\varepsilon$$
Note that $x>1-δ$ and $|x|<1$  means  $|1-x|<δ.$
Thus, it follows that for every $ε>0$ there exists $δ>0$ such that $|1-x|<δ⇒ |f(x)-s|<ε.$
Thus, we conclude that $\lim_{x→1}\;f(x)=s=∑_{n=0}^\infty\;c_n .$
