# How do you find the limit lim (3^(x+5)-2^(2x+1))/(3^(x+1)-2^(2x+4)) as x->oo?

Jan 2, 2018

$\frac{1}{8}$

#### Explanation:

$\frac{{3}^{x + 5} - {2}^{2 x + 1}}{{3}^{x + 1} - {2}^{2 x + 4}}$

Rewriting:

$\frac{{3}^{x + 5} - {4}^{x + \frac{1}{2}}}{{3}^{x + 1} - {4}^{x + 2}}$

Concentrating on the dominating terms of numerator and denominator:

$\frac{- {4}^{x + \frac{1}{2}}}{- {4}^{x + 2}} = \frac{{4}^{x + \frac{1}{2}}}{{4}^{x + 2}}$

Dividing:

$\left({4}^{- \frac{3}{2}}\right) = \frac{1}{{4}^{\frac{3}{2}}} = \frac{1}{8}$

${\lim}_{x \to \infty} \left(\frac{{3}^{x + 5} - {2}^{2 x + 1}}{{3}^{x + 1} - {2}^{2 x + 4}}\right) = \frac{1}{8}$

Jan 2, 2018

$\frac{1}{8}$

#### Explanation:

$\frac{{3}^{x + 5} - {2}^{2 x + 1}}{{3}^{x + 1} - {2}^{2 x + 4}} = \frac{{3}^{5} {3}^{x} - 2 \cdot {4}^{x}}{3 \cdot {3}^{x} - {2}^{4} {4}^{x}}$

$= \frac{{3}^{5} {\left(\frac{3}{4}\right)}^{x} - 2}{3 {\left(\frac{3}{4}\right)}^{x} - 16}$

As $x \rightarrow \infty$, we have ${\left(\frac{3}{4}\right)}^{x} \rightarrow 0$, so the limit sought

is $\frac{0 - 2}{0 - 16} = \frac{1}{8}$