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#