# What is the limit of f(x) = (1 + 4e^-x )/( 3 - 2e^-x) as x goes to infinity?

${\lim}_{x \rightarrow \infty} \frac{1 + 4 {e}^{-} x}{3 - 2 {e}^{-} x} = \frac{1}{3}$
As $x \rightarrow \infty$, we have ${e}^{-} x \rightarrow 0$, so $1 + 4 {e}^{-} x \rightarrow 1$ while $3 - 2 {e}^{-} x \rightarrow 3$.
Interestingly, as $x \rightarrow - \infty$ we get a different limit. We have
${\lim}_{x \rightarrow - \infty} \frac{1 + 4 {e}^{-} x}{3 - 2 {e}^{-} x} = \frac{4}{-} 2 = - 2$