# Limit of a function

hopelesss

## Homework Statement

Hey guys
im trying to figure out what lim x->1 ((x^3)-3x+2) / x-1 is.
I get -1/0 and then when i try factor it i cant get it right..

?

## The Attempt at a Solution

lim x->1 x^3-3x+2 / x-1 = lim x->1 1^3-3*1+2 / 1-1 = -1/0

Mentor

## Homework Statement

Hey guys
im trying to figure out what lim x->1 ((x^3)-3x+2) / x-1 is.
I get -1/0 and then when i try factor it i cant get it right..
x3 -3x + 2 is divisible by x - 1. Do you know how to do polynomial long division. If not, there's an article on this technique on wikipedia.

?

## The Attempt at a Solution

lim x->1 x^3-3x+2 / x-1 = lim x->1 1^3-3*1+2 / 1-1 = -1/0

When you write expressions like the above, put parentheses around the entire numerator and the entire denominator, like so:
(x3 - 3x + 2)/(x - 1)

Without them, what you wrote is x3 - 3x + (2/x) - 1.

hopelesss
I do know abit polynomial division, but i cant see my teacher using this method.
I did mean (x3 - 3x + 2)/(x - 1).

Im used to doing it like this
lim x-> 3 (x-3) / (x3 -9) = lim -> 3 (x-3) / ((x-3)(x+3)) = 1/(3+3) = 1/6
But when i try do this i cant get it right, and i cant cancel the things i dont want 2 have.

clamtrox
Okay, here's an alternative approach: do you remember that you can write a polynomial into a product form like this
$$p(x) = x^3 + a_2 x^2 + a_1 x + a_0 = (x-r_1)(x-r_2)(x-r_3)$$
where r are the roots of the polynomial?

Now you already noticed that one of the roots is +1. Just write out the right hand side and make all of the coefficients equal. You'll get some equations for the remaining roots, and they should be easy to solve.

Mentor
I do know abit polynomial division, but i cant see my teacher using this method.
You're the one working this problem, not your teacher. Besides, you are probably underestimating your teacher's abilities. If you know about polynomial long division, why not use it? (x3 - 3x + 2) factors nicely.
I did mean (x3 - 3x + 2)/(x - 1).

mtayab1994
Use x-1 as a factor in the numerator and then cancel it out with the x-1 in the denominator. BTW if you want to factor with x-1 you'll have to do something called synthetic division.

Staff Emeritus
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