Evaluating limit by factorization

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Discussion Overview

The discussion revolves around evaluating a limit using the factorization method, specifically the limit $\lim_{x\to5} \frac{x^3 + 3x^2 - 6x + 2}{ x^3 + 3x^2 - 3x - 1}$. Participants explore the appropriateness of factorization and the results obtained through direct substitution versus factorization.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant requests assistance in solving the limit using the factorization method, expressing uncertainty about how to factor the polynomial.
  • Another participant questions the need for factorization, suggesting that direct substitution is sufficient since the limit does not appear to be in an indeterminate form.
  • A participant notes a discrepancy between their direct substitution result of 9 and a book answer of -11, prompting further investigation into the problem.
  • Another participant calculates the limit and finds it to be $\frac{43}{46}$, questioning the correctness of the original problem statement.
  • A later reply suggests that the problem may have been misinterpreted, proposing an alternative limit expression that leads to the answer of -11 when evaluated at $x = -5$.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct limit or the necessity of factorization. Multiple competing views remain regarding the interpretation of the problem and the methods of evaluation.

Contextual Notes

There are unresolved issues regarding the correct formulation of the limit problem and the assumptions about the need for factorization versus direct substitution.

Joel Jacon
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Can anyone tell me how to solve the following limit by factorization method
$\lim{{x}\to{5}} \frac{x^3 + 3x^2 - 6x + 2}{ x^3 + 3x^2 - 3x - 1}$?Please tell me how to factorize such big equation?
 
Last edited:
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Why do you want to factorize it?
The factorization method is useful when the limit is of an indeterminate form like $\frac{0}{0}$ or $\frac{\infty}{\infty}$. But this is not the case thus you can just plug in the value $x=5$.
 
But the answer given in my book is -11. While using direct substitution I get 9. How can you get -11
 
Last edited:
$$\lim_{x\to5}\frac{x^3+3x^2-6x+2}{x^3+3x^2-3x-1}=\frac{172}{184}=\frac{43}{46}$$$$\text{ }$$Are you sure you typed the problem correctly?
 
Yes, the question is correct. See the question 1 in the image
 

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After saving the image, and rotating it so that it is not upside down, then straining my eyes to read the out of focus image, what I see is:

1.) $$\lim_{x\to5}\frac{2x^2+9x-5}{x+5}$$

Now, you can factor as follows (although it is not necessary):

$$\lim_{x\to5}\frac{(2x-1)(x+5)}{x+5}=\lim_{x\to5}2x-1=2(5)-1=9$$

Apparently what was meant, if an answer of $-11$ was given is:

$$\lim_{x\to-5}\frac{2x^2+9x-5}{x+5}=2(-5)-1=-11$$
 

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