# What is the limit (x->0) of (sin 3x)/x The answer is 3 but I dont know how to get it.

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Evaluating this limit uses the fundamental trig limit `lim_{x->0}{sin x}/x` and a substitution.

`lim_{x->0}{sin 3x}/x` multiply by 3/3

`=3 lim_{x->0} {sin 3x}/{3x}` change variables let u=3x

`=3lim_{u->0} {sin u}/u` now use limit

`=3(1)`

`=3`

**The limit is 3.**

The limit `lim_(x->0)sin (3x)/x` has to be determined.

If we substitute x = 0 in `sin (3x)/x` , the result obtained is of the form `0/0` . This is an indeterminate number and l"hospital's rule can be used in this case to arrive at the required limit.

Replace the numerator and denominator by their derivatives.

`(sin 3x)' = 3*cos (3x)`

`(x)' = 1`

This gives the limit:

`lim_(x->0) (3*cos (3x))/1`

Now cos 0 = 1. Substituting x = 0 in the expression obtained above gives:

`(3*1)/1 = 3`

The required limit `lim_(x->0)sin (3x)/x = 3`