Question

The value of $\underset{n\to \infty }{lim}\frac{3^{n+1}+2^{n+2}}{{5.3}^n-2^{n-1}}$ is

• A. $\frac{1}{5}$
• B. $\frac{2}{5}$
• C. $\frac{3}{5}$
• D. $1$

Answer 1

The correct option is C $\frac{3}{5}$

Given:
$\underset{n\to \infty }{lim}\frac{3^{n+1}+2^{n+2}}{5\times 3^n-2^{n-1}}$
$\Rightarrow \underset{n\to \infty }{lim}\frac{3^n\times 3+2^n\times 2^2}{5\times 3^n-2^n\times 2^{-1}}$
Taking $3^n$ common, We get
$\Rightarrow \underset{n\to \infty }{lim}\frac{3^n\times [3+2^2\times {\frac{2}{3}}^n]}{3^n\times [5-2^{-1}\times {\frac{2}{3}}^n]}$
$\Rightarrow \underset{n\to \infty }{lim}\frac{[3+2^2\times {\frac{2}{3}}^n]}{[5-2^{-1}\times {\frac{2}{3}}^n]}$
$\because \underset{n\to \infty }{lim}(\frac{2}{3}{)}^n=0$
On further solving, it will reduce to
$\Rightarrow \frac{3}{5}$
Therefore $\underset{n\to \infty }{lim}\frac{3^{n+1}+2^{n+2}}{5\times 3^n-2^{n-1}}=\frac{3}{5}$