Let #f(x) = sec(3x)cos(5x) = cos(5x)/cos(3x)#.
#f(x) = 1# whenever #x# is an integer multiple of #pi#. To show this we can look at cases.
If #x = kpi# for even #k#, then #cos(5x)=cos(3x) = 1#
If #x = kpi# for odd #k#, then #cos(5x)=cos(3x) = -1#
In either case, we have #f(x) = 1#.
Of course this occurs infinitely many times as #x rarroo#.
Every time #x# is an odd multiple of #pi/2# we have one of three cases. One is enough to establish the nonexistence of a limit of #f(x)# as #xrarroo#.
If #cos(5x)=0# and #cos(3x) != 0#, then #f(x) = 0#
This also occurs infinitely many times as #xrarroo#.
As #xrarroo#, #f(x)# hits both #0# and #1# infinitely many times. So, we see that #f(x)# cannot approach a limit.
