lim-n-2-n-4-n-3-2-n-3-4n-

Question Number 131885 by Study last updated on 09/Feb/21

{l i m}_{n \to - \infty} \frac{2^{n} + 4^{n - 3}}{2^{n - 3} + 4 n} = ?

Commented by malwan last updated on 09/Feb/21

\underset{n \to - \infty}{l i m} \frac{2^{n} + 4^{n - 3}}{2^{n - 3} + 4^{n}} = \underset{n \to - \infty}{l i m} \frac{2^{n} + 2^{2 n - 6}}{2^{n - 3} + 2^{2 n}}

= \underset{n \to - \infty}{l i m} \frac{2^{n} (1 + 2^{n - 6})}{2^{n} (2^{- 3} + 2^{n})} = \frac{1}{2^{- 3}} = 8

Answered by EDWIN88 last updated on 10/Feb/21

if \lim_{n \to - \infty} (\frac{2^{n} + 4^{n - 3}}{2^{n - 3} + 4^{n}}) = \lim_{n \to - \infty} (\frac{4^{n} [{(\frac{2}{4})}^{n} + \frac{1}{64}]}{4^{n} [1 + \frac{1}{8} {(\frac{2}{4})}^{n}})

= \lim_{n \to - \infty} (\frac{\frac{1}{2^{n}} + \frac{1}{64}}{1 + \frac{1}{8} . \frac{1}{2^{n}}}) = \lim_{n \to - \infty} (\frac{1 + 64 . 2^{n}}{2^{n} + \frac{1}{8}}) = 8

note that {\underset{n \to - \infty}{\lim 2}}^{n} = 0

Answered by liberty last updated on 10/Feb/21

If \lim_{n \to - \infty} (\frac{2^{n} + 4^{n - 3}}{2^{n - 3} + 4^{n}})

let 2^{n} = x where x \to 0, as n \to - \infty

then \lim_{x \to 0} \frac{x + \frac{1}{64} x^{2}}{\frac{1}{8} x + x^{2}} = \lim_{x \to 0} \frac{64 x + x^{2}}{8 x + {64 x}^{2}}

= \lim_{x \to 0} [\frac{64 + x}{8 + 64 x}] = \frac{64}{8} = 8

Facebook Tweet Pin