Întrebare limita sir

Buna ziua
Se considera functia f care satisface relatia:
\[2f\left(\dfrac{x-2}{x+1}\right)+f\left(\dfrac{x+1}{x-2}\right)=x\\\]
Se cere sa se calculeze limita sirului cu termenul general:
\[a_n=\dfrac{1}{f^{(n)}(-1)},\ unde\ f^{(n)}(x)\]semnifica derivata de ordinul n
a lui f.
Rezultat valabil(unul din ele):
\[a)0;b)\dfrac{2}{3};c)\dfrac{3}{2};d)3;e)\infty\]
multumesc

Facem substitutia \(\frac{x-2}{x+1}=t\), atunci \(\Rightarrow x-2=tx+t\Rightarrow x=\frac{t+2}{1-t}\).

Se poate scrie \(2f\left ( t \right )+f\left ( \frac{1}{t} \right )=\frac{t+2}{1-t}\).
Punand \(t\rightarrow \frac{1}{t}\) : \(2f\left ( \frac{1}{t} \right )+f\left ( t \right )=\frac{\frac{1}{t}+2}{1-\frac{1}{t}}\Leftrightarrow f\left ( t \right )+2f\left ( \frac{1}{t} \right )=\frac{2t+1}{t-1}\).

Am obtinut sistemul \(\left\{\begin{matrix} 2f\left ( t \right )+f\left ( \frac{1}{t} \right )=\frac{t+2}{1-t}\\ f\left ( t \right )+2f\left ( \frac{1}{t} \right )=\frac{2t+1}{t-1} \end{matrix}\right.\).

De aici se obtine \(f\left ( t \right )=\frac{1}{3}\cdot \frac{4t+5}{1-t}\), adica \(f\left ( x \right )=\frac{1}{3}\cdot \frac{4x+5}{1-x}\).

Derivatele functiei \(f\): \(f'\left ( x \right )=3\cdot \frac{1}{\left ( 1-x \right )^{2}}\), \(f''\left ( x \right )=3\cdot \frac{2}{\left ( 1-x \right )^{3}}\), \(f'''\left ( x \right )=3\cdot \frac{2\cdot 3}{\left ( 1-x \right )^{4}}\), ..., \(f^{\left ( n \right )}\left ( x \right )=3\cdot \frac{n!}{\left ( 1-x \right )^{n+1}}\), \(n\in \mathbb{N}^{*}\).

Sirul arata asa: \(a_{n}=\frac{1}{f^{\left ( n \right )}\left ( -1 \right )}=\frac{2^{n+1}}{3n!}=\frac{2}{3}\cdot \frac{2^{n}}{n!}\).

Cum \(\frac{2^{n}}{n!}\rightarrow 0\), cand \(n\rightarrow \infty\), rezulta ca \(a_{n}=\frac{2}{3}\cdot \frac{2^{n}}{n!}\rightarrow 0\), \(n\rightarrow \infty\).

Varianta corecta de raspuns: \(a)\).

Buna seara
Am inteles perfect rezolvarea multumesc!