Find The Limit sequences I tried to calculate this limit But I couldn't find the solution
Is there a hint or someone who has an idea how to calculate it ??
$$\forall\;n\in\;N^{*}$$
$$\underset{n\rightarrow\infty}{lim\;\;inf_{x\in\;R}\;(\sum_{k＝0}^{2n}x^k)}$$
After simplifying I found this :
$$
\underset{n\rightarrow\infty}{lim\;\;inf_{x\in\;R}\;(\frac {x^{2n＋1}-1}{x－1}})$$
I'm a new user on this site, so my questions seem kind of bad to ask

 A: The function $f_n(x)=\sum_{k=0}^{2n}x^k$ satisfies $f_n(x)\ge 1$ for $x\notin (-1,0).$ Indeed it is clear for $x\ge 0.$ For $x\le -1$ we have
$$f_n(x)-1=\sum_{k=0}^{2n}x^{2k}-1=\sum_{k=1}^{2n}x^{2k}={x\over x-1}(x^{2n}-1)\ge 0$$
Therefore the minimal value may occur only in the interval $(-1,0).$
We have
$$f_n'(x)={(2n+1)x^{2n}(x-1)-(x^{2n+1}-1)\over (x-1)^2}={2n x^{2n+1}-(2n+1)x^{2n}+1\over (x-1)^2}$$ Consider
$$g_n(t)=-2n t^{2n+1}-(2n+1)t^{2n}+1,\quad 0<t<1 $$
Observe that $g_n(-x)=f_n'(x)(x-1)^2.$
The function $g_n(t)$ is decreasing in the interval $(0,1)$ and $g_n(0)=1,$ $g_n(1)= -4n.$ Therefore $g_n(t)$ vanishes exactly once at a point $0<t_{n,0}<1$ and changes the sign at $t_{n,0}$ from $+$ to $-.$  Thus $f_n'(x)$ vanishes exactly once at a point $x_{n,0}=-t_{n,0}$ and changes sign at $x_{n,0}$ from $-$ to $+.$ Hence the function $f_n$ attains its minimal value at $x_{n,0}.$
Let $$t_{n,1}^{2n+1}={1\over 4n},\qquad t_{n,2}^{2n}={1\over 2(2n+1)}$$ Then $0<t_{n,1},t_{n,2}<1$ and
$$g_n(t_{n,1})= -{1\over 2}-{2n+1\over 4nt_{n,1}}+1<{1\over 2}-{2n+1\over 4n}<0$$
$$g_n(t_{n,2})= -{n\over 2n+1}t_{n,2}-{1\over 2}+1\ge -{n\over 2n+1}+{1\over 2}>0$$
Hence $t_{n,1}<t_{n,0}<t_{n,2}$ and $t_{n,1}\to 1,\ t_{n,2}\to 1.$
We have
$$f_n(x_{n,0})={x_{n,0}^{2n+1}-1\over x_{n,0}-1}={t_{n,0}^{2n+1}+1\over t_{n,0}+1}$$ Thus
$$ {t_{n,1}^{2n+1}+1\over t_{n,2}+1}\le f_n(x_{n,0})\le {t_{n,2}^{2n+1}+1\over t_{n,1}+1}$$
Furthermore
$${t_{n,1}^{2n+1}+1\over t_{n,2}+1}={{1\over 4n}+1\over t_{n,2}+1}\to {1\over 2}$$
$${t_{n,2}^{2n+1}+1\over t_{n,1}+1}={{1\over 2(2n+1)}t_{n,2}+1\over t_{n,1}+1}\to {1\over 2}$$
Hence $$\lim_nf_n(x_{n,0})={1\over 2}$$
