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May12-07, 10:42 AM
P: 1,874
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)/(5-x) then you would start of with
-1<= sin5x <=1
0 <= sin^2(5x) <=1

then divide by the (5-x) and since x approaches infinity it will be negative and reverse the equalities

0/(5-x) >= sin^2(5x)/(5-x) >= 1/(5-x)

Then the limit as x->inft of 1/(5-x) = 0, so sin^2(5x)/(5-x) = 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