given that the limit $\lim _{n \to \infty} \frac{x_n}{x_{n+1}}=x$ exists, find the value of $x$

Question

suppose $\{x_n\}$ sequence with $x_0=x_1=1$ and $x_n=x_{n-1}+x_{n-2}$ for all natural $n \ge 2$. given that the limit $\lim _{n \to \infty} \frac{x_n}{x_{n+1}}=x$ exists, find the value of $x$

My attempt:

Given that $x_n=x_{n-1}+x_{n-2}$ so the characteristic equation for corresponding this is $x^2-x-1=0$

then $x=\frac{1\pm \sqrt{5}}{2}$

then $x_n=c_1 (\frac{1+\sqrt{5}}{2})^n+c_2 (\frac{1-\sqrt{5}}{2})^n$

But getting $x_n$ difficult this method

Answer 1

Hint: Let $y_n=\dfrac{x_n}{x_{n+1}}$. Then $$ \frac1{y_n} = \frac{x_{n+1}}{x_n} = \frac{x_{n}+x_{n-1}}{x_n} = 1 + \frac{x_{n-1}}{x_n} = 1 + y_{n-1} $$ Now take the limit on both sides.

Answer 2

$\dfrac{X_n}{X_{n+1}} = \cfrac{1}{\cfrac{X_{n+1}}{X_n}}$

Now using the recursive relation, $X_{n+1} = X_n + X_{n-1}$

$\dfrac{X_n}{X_{n+1}} = \cfrac{1}{\cfrac{X_n + X_{n-1}}{X_n}} = \cfrac{1}{1+\cfrac{X_{n-1}}{X_n}} = \cfrac{1}{1+\cfrac{1}{\cfrac{X_{n}}{X_{n-1}}}} $

Now again using recursive relation

$ \cfrac{1}{1+\cfrac{1}{\cfrac{X_{n}}{X_{n-1}}}} = \cfrac{1}{1+\cfrac{1}{\cfrac{X_{n-1}+X_{n-2}}{X_{n-1}}}} = \cfrac{1}{1+\cfrac{1}{1+\cfrac{X_{n-2}}{X_{n-1}}}}$

Now we see the pattern here,if we continue doing like this n times we will get somewhere $1+\cfrac{X_0}{X_1}$

$\lim_{n \to \infty} \dfrac{X_n}{X_{n+1}} = \cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+... }}}}$

$X = \cfrac{1}{1+X}$
$X^2 + X = 1$

$X =\cfrac{√5-1}{2}$