So, I want to prove that the function $ \frac{1}{\left(x-2\right)^{2}} $ is analytic in $ \mathbb{R} \setminus [2] $ What I've tried:
let $ I\in\mathbb{R}\setminus\left\{ 2\right\} $ be open sigment and let $ x_{0}\in I $ . I porved by induction that the nth deriviative satisfies: $ \left(\frac{1}{\left(x-2\right)^{2}}\right)^{(n)}=\left(-1\right)^{n}\frac{\left(n+1\right)!}{\left(x-2\right)^{n+2}} $
now if we'll write the function as its Taylor polymonial + the reminder (lagrange) we have:
$ \frac{1}{\left(x-2\right)^{2}}=\sum_{k=0}^{n}\frac{f^{\left(k\right)}\left(x_{0}\right)}{k!}\left(x-x_{0}\right)^{k}+\frac{f^{\left(n+1\right)}\left(\theta\right)}{\left(n+1\right)!}\left(x-x_{0}\right)^{n+1} $
if we'll prove thath the reminder goes to 0 when n goes to $ \infty $ we will prove the claim.
let $ \delta $ be such $ \left(x_{0}-\delta,x_{0}+\delta\right) \subseteq I $ (we will figure out later what exactly $ \delta $ have to be in order to satisfy the claims)
so, from the last claims we can say:
$ |\frac{f^{\left(n+1\right)}\left(\theta\right)}{\left(n+1\right)!}\left(x-x_{0}\right)^{n+1}|=|\frac{\left(-1\right)^{n+1}\frac{\left(n+2\right)!}{\left(\theta-2\right)^{n+3}}}{\left(n+1\right)!}\left(x-x_{0}\right)^{n+1}|=|\frac{\left(-1\right)^{n+1}\left(n+2\right)}{\left(\theta-2\right)^{n+3}}\left(x-x_{0}\right)^{n+1}|=\frac{\left(n+2\right)|x-x_{0}|^{n+1}}{|\theta-2|^{n+1}\cdot|\theta-2|^{2}}=\frac{n+2}{|\frac{\theta-2}{x-x_{0}}|^{n+1}|\theta-2|^{2}} $
when $ \theta $ is between x and $ x_0 $
so all I have left to prove is that $ |\frac{\theta-2}{x-x_{0}}|>1 $
In that case the reminder will strive to 0 as we want.
but im having trouble choosing the right $ \delta $ in oreder to make sure this will happen.
Any ideas will help, Thanks.
Since $\theta$ is between $x_0$ and $x$, you can assume $$ \theta=(1-t)x_0+t x,t\in(0,1).$$ Let $|x-x_0|<\delta$ for some $\delta<\frac12|2-x_0|$. Then by the triangle inequality, $$ |\theta-2|\ge|2-x_0-t(x-x_0)|\ge|2-x_0|-t|x-x_0|>|2-x_0|-|x-x_0|>|2-x_0|-\delta $$ and hence, for $|x-x_0|<\delta$, $$ \frac{|x-x_{0}|}{|\theta-2|}<\frac{\delta}{|2-x_0|-\delta}<1. $$ So \begin{eqnarray} &&\bigg|\frac{f^{\left(n+1\right)}\left(\theta\right)}{\left(n+1\right)!}\left(x-x_{0}\right)^{n+1}\bigg|\\ &=&\frac{\left(n+2\right)}{|\theta-2|^{2}}\cdot\bigg(\frac{|x-x_{0}|}{|\theta-2|}\bigg)^{n+1}\\ &\le&\frac{\left(n+2\right)}{|\theta-2|^{2}}\cdot\bigg(\frac{\delta}{|2-x_0|-\delta}\bigg)^{n+1} \end{eqnarray} from which, you can derive $$ \lim_{n\to\infty}\bigg|\frac{f^{\left(n+1\right)}\left(\theta\right)}{\left(n+1\right)!}\left(x-x_{0}\right)^{n+1}\bigg|=0. $$