Evaluate #lim _(x-> oo) (sinxcosx)/(3x) # ?

1 Answer
Oct 12, 2016

#lim _(xrarroo) (sinxcosx)/(3x) =0#

Explanation:

Use the Squeeze Theorem at infinity.

Since, #sinx# and #cosx# are between #-1# and #1#, so is their product.

#-1 <= sinxcosx <= 1#

We are interested in #xrarroo#, so we are interested in positive values of #x#.
When #x# is positive, #3x# is positive, so we can divide the inequality without reversing the directions of the inequalities.

#(-1)/(3x) <= (sinxcosx)/(3x) <= 1/(3x)#

#lim_(xrarroo)(-1)/(3x) = 0# and #lim_(xrarroo)(1)/(3x) = 0#.

Therefore, #lim _(xrarroo) (sinxcosx)/(3x) =0#.