Question

$\underset{n\to \infty }{\lim }\frac{3.2^{n+1}-4.5^{n+1}}{5.2^n+7.5^n}$ =

• A. $0$
• B. $\frac{3}{5}$
• C. $\frac{-4}{7}$
• D. $\frac{-20}{7}$

Answer 1

Answer: (D)

$\underset{n\to \infty }{\lim }\frac{3\cdot 2^{n+1}-4\cdot 5^{n+1}}{5\cdot 2^n+7\cdot 5^n}$
$=\underset{n\to \infty }{\lim }\frac{5^n\left(6\cdot {\left(\frac{2}{5}\right)}^n-20\right)}{5^n\left(5\cdot {\left(\frac{2}{5}\right)}^n+7\right)}$
$=-\frac{20}{7}$