Question

Question #6c902

Answer 1

I assume we want `lim_(x->1) ( 3/(x^3-1)-1/(x-1))`. (I forgot to edit the question. )

Explanation:

`lim_(x->1) ( 3/(x^3-1)-1/(x-1)) = lim_(x->1) ( 3/(x^3-1)-(x^2+x+1)/(x^3-1))`

`= lim_(x->1) (-(x^2+x-2))/(x^3-1)`

`= lim_(x->1) (-cancel((x-1))(x+2))/(cancel((x-1))(x^2+x+1))`

`= (-3)/3 = -1`

graph{3/(x^3-1)-1/(x-1) [-3.433, 6.43, -3.157, 1.776]}