# What is the limit as x approaches infinity of e^(-3x) cos(x)?

We have ${\lim}_{x \to \infty} {e}^{- 3 x} \cos \left(x\right) = 0$.

For the limit as $x$ apporaches infinity, we must note that $\cos \left(x\right)$ is a limited function, that is, there exists a number $M$ such that, for every value of $x$, $- M \le q \cos \left(x\right) \le q M$. For $\cos \left(x\right)$, we can take $M = 1$ (or any other value greater than $1$).

Multiplying the inequalities $- 1 \le q \cos \left(x\right) \le q 1$ by ${e}^{- 3 x}$, we get

$- {e}^{- 3 x} \le q {e}^{- 3 x} \cos \left(x\right) \le q {e}^{- 3 x}$

Since ${\lim}_{x \to \infty} - {e}^{- 3 x} = 0 = {\lim}_{x \to \infty} {e}^{- 3 x}$, we have, by the Squeeze Theorem:

${\lim}_{x \to \infty} {e}^{- 3 x} \cos \left(x\right) = 0$