Lim (sin3t)/(sin2t) x-->0 how do we find it's limit?

1 Answer
Jul 12, 2018

lim_(x->0)sin(3t)/sin(2t) = 3/2

Explanation:

Note that:

sin(3t)/sin(2t) = 3/2 sin(3t)/(3t) (2t)/sin(2t)

Consider now the limit:

lim_(x->0) sin(ax)/(ax) with a>0

Substituting y=ax we have that for x->0 als y->0, so:

lim_(x->0) sin(ax)/(ax) = lim_(y->0) siny/y =1

So:

lim_(x->0)sin(3t)/sin(2t) = 3/2 lim_(x->0) sin(3t)/(3t) lim_(x->0) (2t)/sin(2t) =3/2*1*1 = 3/2

graph{sin(3x)/sin(2x) [-10, 10, -5, 5]}#