Question

Punctul c) va rog frumos

Answer 1

$\displaystyle Avem~f'(x)=3x^2+ \frac{1}{3}>0,~deci~f~este~strict~crescatoare. \\ \\ Vom~demonstra~prin~inductie~ca~sirul~dat~este~strict~crescator. \\ \\ Avem~x_1=1+ \frac{1}{3}>1=x_0. \\ \\ Presupunem~x_k<x_{k+1}~pentru~k \in \mathbb{N}~si~demonstram~ca~x_{k+1}<x_{k+2}. \\ \\ Din~f~strict~crescatoare~si~x_k<x_{k+1}~rezulta~f(x_k)<f(x_{k+1}) \Leftrightarrow \\ \\ \Leftrightarrow x_{k+1}<x_{k+2}.$

$\displaystyle Deci~(x_n)_{n \ge 0}~este~strict~crescator.~Prin~urmare~exista~\lim_{n \to \infty}x_n=l \in \overline{\mathbb{R}}. \\ \\ Presupunem~ca~l \in \mathbb{R}.~(adica~l \neq \infty)~Trecand~la~limita~in~relatia \\ \\ x_{n+1}=f(x_n) \Leftrightarrow x_{n+1}=x_n^3+ \frac{x_n}{3},~obtinem~l=l^3+ \frac{l}{3} \Leftrightarrow l^3= \frac{2l}{3} \Leftrightarrow \\ \\ \Leftrightarrow l \left(l^2- \frac{2}{3} \right)=0.$

$\displaystyle Dar~l>x_0=1.~Rezulta~l^2-\frac{2}{3}>0,~deci~l \left(l^2- \frac{2}{3} \right)>0,~contradictie! \\ \\ Asadar~l= \infty.$