Mathematics - Limits Question with Solution | TestHub

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 ${\lim }_{n\to \infty }{x}_n=l$. Therefore

$$\begin{array}{rlrl}l& =\underset{n\to \infty }{\lim }{x}_{n+1}& =\underset{n\to \infty }{\lim }\left(\frac{4+3x_n}{3+2x_n}\right)& =\frac{4+3{\lim }_{n\to \infty }{x}_n}{3+2{\lim }_{n\to \infty }{x}_n}\end{array}$$

$$=\frac{4+3l}{3+2l}$$

Hence $3l+2l^2=4+3l$

or $l^2=2\Rightarrow l=\sqrt{2}\left(\because x_n>0\forall n\right)$.