Evaluate the limit, if it exists. (If an answer does not exist, enter DNE.) limit as x-->0  absvalue(3x-1)−absvalue(3x+1)/ (x)

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You need to evaluate side limits to check if the limit of the function at x=0 exists.

You need to express the absolute values for x>0 and x<0 such that:

`|3x-1| = 3x - 1 if 3x - 1 gt=0 =gt 3x gt=1 =gt xgt=1/3`

`|3x-1| = -3x+ 1 if 3x - 1 lt0 =gt 3x lt1 =gt xlt1/3`

`|3x+1| = 3x+ 1 if 3x+ 1 gt=0 =gt 3x gt=-1 =gt xgt=-1/3 `

`|3x+1| = -3x - 1 lt0 =gt 3x lt1 =gt xlt-1/3`

Notice you need to evaluate the limit for x>0 and x<0, hence you need to work in interval `[-1/3,1/3)`  considering `|3x-1| = -3x+ 1`  and `|3x+1| = -3x- 1`  .

`lim_(x-gt0,xlt0)(-3x + 1 - 3x - 1)/x = lim_(x-gt0,xgt0)(-3x + 1 - 3x - 1)/xlim_(x-gt0,xlt0)(-6x)/x = -6`

Hence, evaluating the limit under the given conditions, `x in [-1/3,1/3),`  yields that the limit exists and it has the value -6.

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