How do you find the limit of #sec3xcos5x# as x approaches infinity?

Question

Scarlett Allison · Accepted Answer

There is no limit. (The limit does not exist.)

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.

Emily Ammerman · Answer

undefined To find the limit of sec^3(x)cos(5x) as x approaches infinity, we can use the properties of limits and trigonometric identities.

First, we note that as x approaches infinity, sec^3(x) will approach infinity since sec(x) approaches infinity as x approaches infinity.

Next, we consider the behavior of cos(5x) as x approaches infinity. Since the cosine function oscillates between -1 and 1, it does not have a limit as x approaches infinity.

Therefore, the limit of sec^3(x)cos(5x) as x approaches infinity does not exist.