Find the Limit of a Function: x->1 ((x^3)-3x+2) / x-1

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Homework Help Overview

The problem involves finding the limit of the function ((x^3)-3x+2) / (x-1) as x approaches 1. Participants are exploring the behavior of the function at this point, particularly focusing on the indeterminate form encountered.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss polynomial long division as a method for simplifying the expression. There are questions about the correct application of factoring and the use of parentheses in expressions. Some participants share their experiences with similar limit problems to draw parallels.

Discussion Status

There is an ongoing exploration of different approaches to the problem, including polynomial division and factoring. Some participants have offered guidance on techniques such as synthetic division, while others are clarifying misunderstandings about the function's evaluation at x=1.

Contextual Notes

Participants note the importance of correctly interpreting the limit and the expression's setup, with some confusion regarding the evaluation leading to an indeterminate form. There is also mention of constraints related to homework rules and expectations for problem-solving methods.

hopelesss
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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 can't get it right..

Homework Equations



?

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
 
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hopelesss said:

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 can't 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.
hopelesss said:

Homework Equations



?

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.
 
I do know abit polynomial division, but i can't 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 can't get it right, and i can't cancel the things i don't want 2 have.
 
Okay, here's an alternative approach: do you remember that you can write a polynomial into a product form like this
[tex]p(x) = x^3 + a_2 x^2 + a_1 x + a_0 = (x-r_1)(x-r_2)(x-r_3)[/tex]
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.
 
hopelesss said:
I do know abit polynomial division, but i can't 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.
hopelesss said:
I did mean (x3 - 3x + 2)/(x - 1).
 
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.
 
hopelesss said:
lim x->1 1^3-3*1+2 / 1-1 = -1/0
You do realize that ##1^3-3\times 1 + 2## is equal to 0, not -1, right?
 
vela said:
You do realize that ##1^3-3\times 1 + 2## is equal to 0, not -1, right?

HAHAHAHA :D no clue what I've been doing. thx!
 

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