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

${e}^{- 3 x} = \frac{1}{{e}^{3 x}}$
As $x \rightarrow \infty$, the denominator ${e}^{3 x} \rightarrow \infty$, so the expression:
${e}^{- 3 x} = \frac{1}{{e}^{3 x}} \rightarrow 0$