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

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Answer 1 · 2018-07-12T13:19:43

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

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]}#