Question

How do I solve for the limit of x->0, xsin(1/16pix)

Answer 1

limx→0x.sin(pix/16)
into quotient in order to use L'Hopital's Rule
limx→0sin(pix/16)/1/x
now differentiate numerator and denominator separately , you get:
limx→0(π/16)cos(pix/16)/1/1
now you can apply the limit, as x approaches 0, y =
π/16