# What is the horizontal asymptote of y=3e^(-1/x) ?

## I plugged in positive and negative infinity into x and got $3 {e}^{0}$ which gave me a horizontal asymptote of 3, is this correct?

$y = 3$
I agree that $y = 3$ is the horizontal asymptote.
Since $- \frac{1}{x}$ approaches 0 as $x$ tends toward both positive and negative infinity, $3 {e}^{- \frac{1}{x}}$ tends toward $3 {e}^{0} = 3$.