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As you can see from the graph below

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

You can also calculate this by using the fact that `lim_(x->0)(sin x)/(x)=1`

We make substitution `t=3x`

`lim_(x->0)(sin 3x)/(x)=lim_(t->0)(sin t)/(t/3)=lim_(t->0)3\cdot(sin t)/t=3lim_(t->0)(sin t)/t=3\cdot1=3`

The alternative is to use L'Hospital's rule.

Given ,

`lim_(x->0) ((sin3x)/x)`

As when value of x is substituted in `(sin3x)/x ` it is of the form 0/0

so , by **using L'Hopital's Rule ,**we get

`lim_(x->0) ((sin3x)/x) =lim_(x->0) ((sin3x)^'/(x^'))`

`=lim_(x->0) ((3*cos3x)/1)`

`= 3*lim_(x->0) (cos3x)`

substituting x --> 0 we get

=`3* (cos(3*0))`

=`3* (cos(0))`

=` 3 ` [since `cos(0) = 1` ]