Question

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

Answer 1

We have `lim_(x to oo) e^(-3x)cos(x) = 0`.

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

Multiplying the inequalities `-1 leq cos(x) leq 1` by `e^(-3x)`, we get

`-e^(-3x) leq e^(-3x)cos(x) leq e^(-3x)`

Since `lim_(x to oo) -e^(-3x) = 0 =lim_(x to oo) e^(-3x)`, we have, by the Squeeze Theorem:

`lim_(x to oo) e^(-3x)cos(x) = 0`