Find a when lim f(x)=(x^2+x+a)/(x-1) = 3 x->1

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Discussion Overview

The discussion revolves around determining the value of \( a \) in the limit expression \( \lim_{x \to 1} \frac{x^2 + x + a}{x - 1} = 3 \). Participants explore different mathematical approaches and reasoning related to limits, continuity, and the behavior of rational functions as they approach points of indeterminacy.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant suggests that since the denominator approaches zero as \( x \to 1 \), the numerator must also approach zero, implying that \( x = 1 \) is a root of the numerator.
  • Another participant agrees that if the limit is to be finite, the numerator must equal zero when \( x = 1 \), leading to the conclusion that \( a = -2 \).
  • A different perspective is presented, arguing that one cannot simply separate the limits of the numerator and denominator without considering their combined behavior, emphasizing the need for \( a = -2 \) to avoid an indeterminate form.
  • Some participants discuss the validity of their reasoning, with one expressing a desire for a theorem to support their intuition about limits.

Areas of Agreement / Disagreement

Participants express differing views on the rigor of separating the limits of the numerator and denominator. While some agree on the necessity of \( a = -2 \) for the limit to be defined, the discussion remains unresolved regarding the best approach to justify this conclusion.

Contextual Notes

There is a lack of consensus on the method of handling limits involving indeterminate forms, particularly regarding the separation of limits and the assumptions made about continuity and reducibility of the function.

jgthb
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Hi

I just found an old high school math exercise and was wondering how it should be answered mathematically correct. The exercise is:

Given lim f(x)=(x^2+x+a)/(x-1) = 3 for x->1, determine a. Back in high school I used to say something like this:

Since the denominator goes to 0 for x -> 1 the fraction must be reducible, so x=1 must be a root in the numerator, and thus we can write it as (x-1)(x-b). Hence:

f(x)=(x-1)(x-b)/(x-1)=x-b.

For x->1 we get:

1-b=3 => b=-2, and thus:

(x-1)(x+2)=x^2+x-2,

so a=-2.

The part I think lacks rigour is the argument that the fraction must be reducible. I guess you have to go on about it in some other way. Any suggestions?

/Jonathan
 
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Welcome to PF!

Hi Jonathan! Welcome to PF! :smile:

(try using the X2 tag just above the Reply box :wink:)
jgthb said:
The part I think lacks rigour is the argument that the fraction must be reducible. I guess you have to go on about it in some other way. Any suggestions?

Looks ok to me … the top must be 0 at x = 1, or the whole thing will be infinite.

But a much quicker way to get a is that if x = 1, the top must be zero, so a = -2.

(More interesting would be to do it for limx->1(x2 + bx + a)/(x-1) = 3, find a and b :wink:)
 
thanks for the reply.

The problem as I see it is that you cannot look at the limit of the numerator and denominator separately, so to me it seems like these arguments doesn't work. But I think i have figured it out:

If you do the following:

lim((x^2+x+a)/(x-1))=lim((x^2+x+a-2+2)/(x-1))=lim((x-1)(x+2)+(a+2))/(x-1))
=lim((x+2)+(a+2)/(x-1))=lim((x+2))+lim((a+2)/(x-1))=3+lim((a+2)/(x-1)).

Now if a != -2 the limit is indefinite (since it goes towards +infinity for x -> 1+ and -infinity for x -> 1-), and thus the limit is only valid for a=-2.
 
jgthb said:
The problem as I see it is that you cannot look at the limit of the numerator and denominator separately, so to me it seems like these arguments doesn't work. But I think i have figured it out:

If you do the following …

Yes, that also works.

But you can look at the limit of the numerator and denominator separately, in that you can say that if the bottom limit is 0, then the top limit must be also, or the result will be ∞.

(in fact, I think that's so obvious, i wouldn't even bother to say it)
 
when it comes to limits I like to have some theorem to backup intuition. but okay, this time it really is obvious as you say, so I'll just accept it :)
 

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