Mathematics - Limits Question with Solution | TestHub

Let ${\mathrm{Lt}}_{n\to \infty }{f}_n=l\Rightarrow {\mathrm{Lt}}_{x\to \infty }f(n+1)=l$.

$$\begin{array}{rl}& {\mathrm{Lt}}_{n\to \infty }f(n+1)=\frac{1}{2}{\mathrm{Lt}}_{n\to \infty }\left[f(n)+\frac{9}{f(n)}\right]\\ & \Rightarrow l=\frac{1}{2}\left[l+\frac{9}{l}\right]\\ & 2l=\frac{l^2+9}{l}\\ & \Rightarrow 2l^2=l^2+9\\ & \Rightarrow l^2=9\\ & \Rightarrow l=3\\ & \because f(n)>0\forall n\in \mathbb{N}\\ & \therefore {\mathrm{Lt}}_{x\to \infty }f(n)=3\end{array}$$