Oh, okay.
So the key to squeeze principle with trig functions is to start off with the basic trig inequalities such as
1<= sin n <=1
and then add on to it
for example if you wanted to find the lim as n>infty of sin^2(5x)/(5x) then you would start of with
1<= sin5x <=1
0 <= sin^2(5x) <=1
then divide by the (5x) and since x approaches infinity it will be negative and reverse the equalities
0/(5x) >= sin^2(5x)/(5x) >= 1/(5x)
Then the limit as x>inft of 1/(5x) = 0, so sin^2(5x)/(5x) = 0.
Your problem is sort of similar, but has a multiplication instead of a division. Start off with
1 <= sin(1/n) <= 1 (there is a discontinuity at n=0, but that is expected since we want to know that limit)
then build
