Find the limit algebraically lim ((x-1)/(x^2-1)) x ->

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Discussion Overview

The discussion revolves around finding the limit of the expression \(\lim_{x \to 1} \frac{x-1}{x^2-1}\) algebraically. Participants explore different methods and interpretations of the problem, focusing on algebraic manipulation versus the application of L'Hôpital's Rule.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant suggests that directly substituting \(x = 1\) leads to an indeterminate form (0/0) and proposes using L'Hôpital's Rule to evaluate the limit, arriving at a result of 1/2.
  • Another participant challenges the use of L'Hôpital's Rule, emphasizing that the question specifically asks for an algebraic solution to the limit.
  • A subsequent reply provides a factorization of the denominator, stating that \(x^2 - 1 = (x-1)(x+1)\), and simplifies the expression to find the limit as \(\frac{1}{2}\) by canceling the common factor of \(x-1\).

Areas of Agreement / Disagreement

There is disagreement regarding the method of finding the limit. Some participants advocate for the use of L'Hôpital's Rule, while others insist on an algebraic approach. The discussion remains unresolved as to which method is preferred or correct.

Contextual Notes

Participants express differing interpretations of what constitutes an algebraic solution, leading to confusion about the requirements of the problem. The discussion highlights the importance of definitions and methods in evaluating limits.

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Find the limit algebraically lim ((x-1)/(x^2-1)) x -->

Find the limit algebreically:

lim ((x-1)/(x^2-1))
x --> 1

Thanks
 
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Hello.

To find your limit we cannot just set x equal to 1. This would make the expression indeterminate, i.e. 0/0. Hence we have to use L ' Hopital's Rule.

lim f ' (x) / g ' (x) = 1 / 2x. Setting x = 1 we get 1/2 for the limit.
x-->1

If you want to solve it algebraically:

lim (x-1)/((x^2 - 1) = lim (x-1)/(x-1)(x+1) = lim 1 / (x+1) = 1/2
x-->1 x-->1 x-->1
 
Last edited:
That's not doing it algebraically. The questions wants you to do something algebraic with (x-1)/(x^2-1)
 
Algebra is the real key here. Factor the denominator

[tex]x^2-1=(x-1)(x+1)[/tex]

Then we have

[tex]\lim\limits_{x \to 1} \frac {x-1}{x^2-1} = \lim\limits_{x \to 1} \frac {x-1}{(x-1)(x+1)} = \lim\limits_{x \to 1} \frac {1}{x+1} = \frac {1}{2}[/tex]
 

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