Question

$\underset{n\to \infty }{lim}\frac{3^{n+1}+2^{n+2}}{3^{n-1}+2^{n-2}}$ is equal to

• A. $4$
• B. $\frac{5}{2}$
• C. $9$
• D. $8$

Answer 1

The correct option is C $9$

$\underset{n\to \infty }{lim}\frac{3^{n+1}+2^{n+2}}{3^{n-1}+2^{n-2}}=\underset{n\to \infty }{lim}\frac{3\cdot 3^n+4\cdot 2^n}{\frac{1}{3}\cdot 3^n+\frac{1}{4}\cdot 2^n}$

Now divide both denominator and numerator by $3^n$
$=\underset{n\to \infty }{lim}\frac{3+4\cdot {\left(\frac{2}{3}\right)}^n}{\frac{1}{3}+\frac{1}{4}\cdot {\left(\frac{2}{3}\right)}^n}=\frac{3+0}{\frac{1}{3}+0}=9$