# lim [sin 3x*tan 2x]/(x-x^3)^2 as x approaches 0 if i use sinx/x=1 -----lim [sin 3x* 3*tan2x*2] / (3x*2x *( x-x^3)^2) what to do with ( x-x^3)^2 ????

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You need to perform the multiplications as you have already done it and you need to continue to group the terms conveniently such that:

`lim_(x-gt0) (sin(3x)/(3x))*(tan(2x)/(2x))*((3x*2x)/((x-x^3)^2))`

Notice that you need to substitute by 1 each special limit `lim_(x-gt0) (sin(3x)/(3x))` and `lim_(x-gt0)(tan(2x)/(2x))` and you need to evaluate `lim_(x-gt0) (3x*2x)/((x-x^3)^2).`

You need to substitute 0 for x in `lim_(x-gt0) (3x*2x)/((x-x^3)^2) ` such that:

`lim_(x-gt0) (3x*2x)/((x-x^3)^2) = (3*0*2*0)/((0-0^3)^2) = 0/0`

Since the evaluation of `lim_(x-gt0) (3x*2x)/((x-x^3)^2)` yields the indeterminate case `0/0` , you need to use l'Hospital's theorem such that:

`lim_(x-gt0) (6x^2)/((x-x^3)^2) = lim_(x-gt0) ((6x^2)')/(((x-x^3)^2)')`

`lim_(x-gt0) ((6x^2)')/(((x-x^3)^2)') = lim_(x-gt0) (12x)/(2*(1-3x^2)(x-x^3))`

You need to substitute 0 for x in `lim_(x-gt0) (12x)/(2*(1-3x^2)(x-x^3))` such that:

`lim_(x-gt0) (12x)/(2*(1-3x^2)(x-x^3)) = 0/0`

You need to use l'Hospital's theorem such that:

`lim_(x-gt0) (12x)/(2*(1-3x^2)((x-x^3)^2)) = lim_(x-gt0) ((12x)')/((2*(1-3x^2)(x-x^3))') `

`lim_(x-gt0) ((12x)')/((2*(1-3x^2)(x-x^3))') = lim_(x-gt0) ((12x)')/(((2-6x^2)(x-x^3))')`

`lim_(x-gt0) ((12x)')/(((2-6x^2)(x-x^3))') =lim_(x-gt0) 12/(12x(x-x^3) + (2-6x^2)(1-3x^2)) = 12/2 = 6`

**Hence, evaluating the limit of the function using special limits and l'Hospital's theorem yields 6.**

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