Question

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 (When $n\in N$)

• A. $[2,5)$
• B. $(1,5)$
• C. $(-1,5)$
• D. $(-\infty ,\infty )$

Answer 1

The correct option is C $(-1,5)$

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

We know $\underset{n\to \infty }{lim}\frac{1}{n}=0$

then we want $\frac{|x-2|}{3}<1$ $\to \because$ if $a<1,a^n=0$ for $n\to \infty$

$x<5$

$x>-1$

$\therefore (-1,5)$.