19: There is just one trouble, at x = 0. The boundedness of sine suggest
that we use the absolute value version of the Comparison test. There is a
natural upper bound
Unfortunately, the corresponding test integral
is divergent,
which means that the test fails.
The Limit Comparison test may help here if you recall that for x close
to 0 we have
sin(x) ∼ x.
is divergent,
which means that the test fails.