Equality of algebaric fractions
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1 Answer

1 | 1 1 0  -2

     1 1 2 | 2

     1 2 2 | 0 = x2+2x+2

x3+x2-2=(x-1)(x2+2x+2), therefore:

(x3+x2-2)/(x-1) simplifies to x2+2x+2 when x≠1, which is the domain. 

Limit as x→1 is 1+2+2=5 although the point (1,5) is undefined for the original expression.

For most intents and purposes, the expression x2+2x+2 is a valid substitute and the domain is unbounded. (The range is restricted to [1,∞).)

by Top Rated User (1.2m points)

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