limit

The way to look at this is that sin(1/x) always lies between -1 and 1 and so as x approaches zero, 0<=|x sin(1/x)| <= |x| and since |x| approaches zero, the smaller thing, |x sin(1/x)| must approach zero also.

also
We know from the range of sin(1/x)
-1 <= sin(1/x) <= 1

multiplying through by x, you obtain
-x <= x*sin(1/x) <= x

taking the limit as x->0 we get
0 <= lim x->0 x*sin(1/x) <= 0

therefore by squeeze/sandwich theorem the
lim x->0 x*sin(1/x) =0

-x <= x*sin(1/x) <= x for x>=0
-x > x*sin(1/x) > x for x< 0

Then right and left handed limits
0 <= lim x->0+ x*sin(1/x) <= 0
0 > lim x->0- x*sin(1/x) > 0