Given that the function
$f(x) = \frac {x^3 +1}{3}$ has three fixed points $ α, β, γ $ in $(−2, −1)$, $(0, 1)$ and$ (1, 2)$ respectively, let us define a sequence of real numbers {$x_n$} as $x_1 = γ − 0.01$ , $x_{n+1} = f(x_n)$,$ n = 1, 2, 3, . .$
Given that the sequence converges, find $\lim_{n→∞}x_n$?
My solution : Here $\{x_n\}$ is monotone, so it converges to a finite $L$
Now if $x_n→L$ then $f(x_n)→f(L)$ because f is continuous but $f(x_n)=x_{n+1}$. Hence $f(L)=L$.
Now $\lim_{n→∞}x_n =\lim_{n→∞}f(L)$,
$f(L) = \frac {L^3 +1} {3}$
$L^3 +1 = 3L$,
$L(L^2-3) = 1 .-1$, that is $L= 1$ and $L^2 -3 = -1$.
ANSWER : $\lim_{n→∞}x_n=L =1$ or $\lim_{n→∞}x_n=L = \sqrt2$.
I'm confused that I'm getting two values of $\lim_{n→∞}x_n$.
Now My question is that what is the correct value of $\lim_{n→∞}x_n$ ????
How did you get these values for the solutions of the equation $f(x)=x$? None of them is correct. What you were supposed to prove was that $\lim_{n\in\mathbb N}x_n=\gamma$. Since the sequence is increasing and $\alpha<\beta<x_1<\gamma$, the limit can only be $\gamma$. In fact, since $x_1\leqslant x_2\leqslant x_3\leqslant\cdots$, the limit $\lim_{n\to\infty}x_n$ cannot be $\alpha$ or $\beta$. Since the limit exists, it can only be $\gamma$.