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tiburtius | High School Teacher | (Level 2) Educator

Posted on

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.

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kspcr111 | In Training Educator

Posted on

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` ]

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