Mathematics - Limits Question with Solution | TestHub

Let $\underset{n\to \infty }{\mathrm{Lt}}fn=l\Rightarrow \underset{x\to \infty }{\mathrm{Lt}}f(n+1)=l$

$$\begin{array}{rlrl}& {\mathrm{Lt}}_{n\to \infty }f(n+1)=\frac{1}{2}\underset{n\to \infty }{\mathrm{Lt}}\left[f(n)+\frac{9}{f(n)}\right]& & \end{array}$$

$$\Rightarrow l=\frac{1}{2}\left[l+\frac{9}{l}\right]2l=\frac{l^2+9}{l}\Rightarrow 2l^2=l^2+9\Rightarrow l^2=9$$

$$l=3$$

$$\begin{array}{rlr}& \because f(n)>0\forall n\in N& \therefore {\mathrm{Lt}}_{x\to \infty }f(n)=3\end{array}$$