# How do you find the limit of  (2^x-3^-x)/(2^x+3^-x) as x approaches infinity?

Jun 22, 2016

1

#### Explanation:

$L = {\lim}_{x \setminus \to \setminus \infty} \frac{{2}^{x} - {3}^{-} x}{{2}^{x} + {3}^{-} x}$

$= {\lim}_{x \setminus \to \setminus \infty} \frac{{2}^{x} - {3}^{-} x}{{2}^{x} + {3}^{-} x} \cdot \frac{{2}^{- x}}{{2}^{- x}}$

$= {\lim}_{x \setminus \to \setminus \infty} \frac{1 - {3}^{- x} \cdot {2}^{- x}}{1 + {3}^{- x} \cdot {2}^{- x}}$

$= {\lim}_{x \setminus \to \setminus \infty} \frac{1 - {6}^{- x}}{1 + {6}^{- x}}$

${\lim}_{x \setminus \to \infty} {6}^{- x} = 0 \setminus \implies L = 1$