Question

How do you find the limit of `(sin 2x)/(sin 3x)` as x approaches 0?

Answer 1

`L_(x->0)(sin2x)/(sin3x)=2/3`

Explanation:

We know that `L_(theta->0)(sintheta)/theta=1`

Hence `L_(x->0)(sin2x)/(sin3x)`

= `L_(x->0)[(sin2x)/(2x)((3x)/sin(3x))xx2/3]`

= `(L_(x->0)(sin2x)/(2x))/(L_(x->0)(sin3x)/(3x))xx2/3`

= `1/1xx2/3=2/3`