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.