Question

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

Answer 1

substitue 0 directly into the limit

Answer 2

change \[\lim_{x \rightarrow 0} x.\sin(pix/16)\] into quotient in order to use L'Hopital's Rule
\[\lim_{x \rightarrow 0} \sin(pix/16)/1/x\]
now differentiate numerator and denominator separately , you get:
\[\lim_{x \rightarrow 0} (\pi/16)\cos(pix/16)/1/1\]
now you can apply the limit, as x approaches 0, y = \[\pi/16\]