# How do you show that lim_(x->\infty)(2^(x+3)-2*3^(x-1))/(2^(x-1)+3^(x-2))=-6?

## How do you show that ${\lim}_{x \to \setminus \infty} \frac{{2}^{x + 3} - 2 \cdot {3}^{x - 1}}{{2}^{x - 1} + {3}^{x - 2}} = - 6$?

Dec 19, 2017

See below.

#### Explanation:

$\frac{{2}^{x + 3} - 2 \cdot {3}^{x - 1}}{{2}^{x - 1} + {3}^{x - 2}} = {3}^{x} / {3}^{x} \left(\frac{{2}^{3} {\left(\frac{2}{3}\right)}^{x} - \frac{2}{3}}{\frac{1}{2} {\left(\frac{2}{3}\right)}^{x} + \frac{1}{9}}\right) = \frac{{2}^{3} {\left(\frac{2}{3}\right)}^{x} - \frac{2}{3}}{\frac{1}{2} {\left(\frac{2}{3}\right)}^{x} + \frac{1}{9}}$

Now

${\lim}_{x \to \infty} \frac{{2}^{x + 3} - 2 \cdot {3}^{x - 1}}{{2}^{x - 1} + {3}^{x - 2}} = \frac{{\lim}_{x \to \infty} {2}^{3} {\left(\frac{2}{3}\right)}^{x} - \frac{2}{3}}{\frac{1}{2} {\lim}_{x \to \infty} {\left(\frac{2}{3}\right)}^{x} + \frac{1}{9}} = \frac{- \frac{2}{3}}{\frac{1}{9}} = - 6$