# lim [ (1+x+x^2)^(1/3) -1] / sin(3x) ?????

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### 1 Answer

The problem does not provide the value x approaches to, hence I suppose that you need to evaluate the limit of function `[(1+x+x^2)^(1/3) -1] / sin(3x)` if x approaches to 0.

`lim_(x-gt0) [root(3)(1+x+x^2) -1] / (sin(3x)) `

You need to substitute 0 for x such that:

`lim_(x-gt0) [root(3)(1+x+x^2) -1] /(sin(3x)) = (root(3)(1) -1)/(sin 0)`

`lim_(x-gt0) [root(3)(1+x+x^2) -1] /(sin(3x)) = 0/0`

Since the evaluation yields an indetermination, hence, the l'Hospital's rule may help you such that:

`lim_(x-gt0) [root(3)(1+x+x^2) -1] /(sin(3x)) = lim_(x-gt0) {[root(3)(1+x+x^2) -1]'} /((sin(3x))')`

`lim_(x-gt0) [(2x+1)/(root(3)((1+x+x^2)^2))] /(3cos(3x))`

Substituting 0 for x yields:

`lim_(x-gt0) [(2x+1)/(root(3)((1+x+x^2)^2))] /(3cos(3x)) = (2*0+1)/(3cos(3*0)*root(3)((1+0+0^2)^2))`

`lim_(x-gt0) [(2x+1)/(root(3)((1+x+x^2)^2))] /(3cos(3x)) = 1/3`

**Hence, evaluating the limit of the function, considering that x approaches to 0, yields `1/3` .**