Question

$\underset{n\to 0}{\lim }\frac{4^{3n-2}-9^{n+1}}{8^{2n-1}-9^{n-1}}=$

• A. $\frac{1}{2}$
• B. 81
• C. Does not exist
• D. None of these

Answer 1

Answer: (C)

$\underset{n\to \infty }{\lim }\frac{4^{3n-2}-9^{n+1}}{8^{2n-1}-9^{n-1}}$
$=\underset{n\to +\infty }{\lim }\frac{4^{-2}\cdot 64^n-9\cdot 9^n}{8^{-1}\cdot 64^n-9^{-1}\cdot 9^n}$
$=\underset{n\to +\infty }{\lim }\frac{4^{-2}-9{\left(\frac{9}{64}\right)}^n}{8^{-1}-9^{-1}{\left(\frac{9}{64}\right)}^n}$
$=\frac{4^{-2}-0}{8^{-1}-0}=\frac{1}{2}$
$=\underset{n\to -\infty }{\lim }\frac{4^{-2}{\left(\frac{64}{9}\right)}^n-9}{8^{-1}{\left(\frac{64}{9}\right)}^n-9^{-1}}$
$=\frac{0-9}{0-9^{-1}}=81$
Hence, limit does not exist.