If $\underset{n\to \infty }{\lim }\frac{n{.3}^n}{n{(x-2)}^n+n{.3}^{n+1}-3^n}=\frac{1}{3},$ then the range of $x$ is (where $n\in N$ ):

We have, $\underset{x\to \infty }{\lim }\frac{n{.3}^n}{n{(x-2)}^n+n{.3}^{n+1}-3^n}=\frac{1}{3}$

Dividing numerator and denominator by $n\times 3^n,$ then $\underset{n\to \infty }{\lim }\frac{1}{{\left(\frac{x-2}{3}\right)}^n+3-\frac{1}{n}}=\frac{1}{3}$

$\Rightarrow \underset{n\to \infty }{\lim }\frac{1}{n}\to 0$ (which is true) and $\underset{n\to \infty }{\lim }{\left(\frac{x-2}{3}\right)}^n\to 0\Rightarrow 0\le \frac{x-2}{3}<1\Rightarrow 2\le x<5$

$\therefore x\in [2,5)$