Question

Let $x_1=1and{x}_{n+1}=\frac{4+3x_n}{3+2x_n}forn\le 1$. If $\underset{n\to \infty }{lim}{x}_n$ exists finitely, then the limit is equal to

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

We have $x_1=1,x_2=\frac{4+3}{3+2}=\frac{7}{5}$
$x_3=\frac{4+3x_2}{3+2x_2}=\frac{4+3\left(\frac{7}{5}\right)}{3+2\left(\frac{7}{5}\right)}=\frac{41}{29}>x_2$
We can easily verify that $x_n<x_{n+1}$ and hence $\left\{x_n\right\}$ is strictly increasing sequence of positive terms. Let $\underset{n\to \infty }{lim}{x}_n=l$. Therefore
$l=\underset{n\to \infty }{lim}{x}_{n+1}$
=$\underset{n\to \infty }{lim}\left(\frac{4+3x_n}{3+2x_n}\right)$
=$\frac{4+3\underset{n\to \infty }{lim}{x}_n}{3+2\underset{n\to \infty }{lim}{x}_n}$
=$\frac{4+3l}{3+2l}$
Hence $3l+2l^2=4+3l$
or $l^2=2\varphi l=\sqrt{2}(\because x_n>0^{\prime \prime }n)$