Evaluate `lim_(x to0 ) ("sin" 3x)/("sin" 6x)`

Explanation

We have, `underset(x to 0)(lim) (sin 3x)/(sin 6x) = underset(x to 0)(lim) [(sin 3x)/(3x) (3x)/(sin 6x) (2)/(3)]`
`= underset(x to 0)(lim) (1)/(2) (sin^(3x))/(3x) underset(x to 0)(lim) (6x)/(sin 6x)`
`= (1)/(2) underset(x to 0)(lim) (sin 3x)/(3x) // underset(x to 0)(lim) (sin 6x)/(6x) = (1)/(2) (1)//(1) (1)/(2) = (1)/(2) [As x to 0 implies 33x to 0 "and" 6x to 0]`