For every integer #N>0# consider points:
#x_N = Npi-pi/2#
they divide the interval #[pi/2,+oo)# in a set of contiguos intervals such that every point #x> pi/2# falls in one of the intervals:
# I_N= [x_N, x_(N+1)) = [Npi-pi/2, Npi+pi/2)#
at the boundaries of every interval we have:
For #N# even:
#sin (x_N) = sin(2Kpi-pi/2) = sin(-pi/2) = -1 #
#sin (x_(N+1)) = sin(2Kpi-pi/2+pi) = sin(pi/2) = 1 #
For #N# odd:
#sin(x_N) = sin((2K+1)pi-pi/2) = sin(pi/2) = 1 #
#sin (x_(N+1)) = sin((2K+1)pi-pi/2+pi) = sin((3pi)/2) = -1 #
In either case for the theorem of intermediate values, as #f(x) = xsinx# is continuous in the intervals #I_N#, it assumes in the interval all the possible values between #-x_N# and #x_N#.
This means that given any number #L# and any numbers #M,Q > 0# we can find a point #bar x > M# such that:
#abs (f(bar x) - L) > Q#
which is in contradiction with #f(x)# converging to any limit finite or infinite.
graph{xsinx [-191.3, 212.9, -100.5, 101.6]}
