First notice that the denominator is zero if we substitute x = -1, showing that the function is not defined when x = -1. Thus, we need to simplify the function algebraically:

(x3+1)(x-1) x2 + 3x + 2	=	(x+1)(x2-x+1)(x-1) (x+1)(x+2)
	=	(x2-x+1)(x-1) (x+2)

Since we are now left with a closed form function that is defined when x = -1, we can now evaluate the limit by substitution:

	lim  x-1	(x3+1)(x-1) x2 + 3x + 2
=	lim  x-2	(x2-x+1)(x-1) (x+2)

=	((-1)2-(-1)+1)((-1)-1) ((-1)+2)
=	-6.

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