Solving Absolute Value Limit: x→2

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Homework Help Overview

The problem involves evaluating the limit as x approaches 2 for the expression involving absolute values: ##\displaystyle \lim_{x\rightarrow 2} \frac{\left | x^2 + 3x + 2 \right |}{x^2 - 4}##. Participants are exploring the implications of the absolute value in the numerator and the behavior of the denominator as it approaches zero.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the factorization of the numerator and the implications of the absolute value. There is uncertainty about how to handle the absolute value when x is near 2. Some participants suggest that since x is close to 2, the expression inside the absolute value will be positive, while others question the correctness of the original problem's numerator.

Discussion Status

The discussion is ongoing, with participants sharing their reasoning and questioning the assumptions made in the problem setup. Some have proposed alternative forms of the limit that may be more straightforward to evaluate, but there is no consensus on a final approach yet.

Contextual Notes

There is a mention of a potential misstatement in the problem regarding the numerator, with some participants suggesting it might be ##x^2 - 3x + 2## instead. The implications of this change on the limit are being explored.

Mr Davis 97
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Homework Statement


Solve: ##\displaystyle \lim_{x\rightarrow 2} \frac{\left | x^2 + 3x + 2 \right |}{x^2 - 4}##.

Homework Equations

The Attempt at a Solution


I am trying to solve the following limit: ##\displaystyle \lim_{x\rightarrow 2} \frac{\left | x^2 + 3x + 2 \right |}{x^2 - 4} = \lim_{x\rightarrow 2} \frac{\left | (x + 1)(x + 2) \right |}{(x - 2)(x + 2)}##. However, I am not sure how to proceed. Normally, I would cancel out the factors, but I am not sure what do with the absolute value in the numerator.
 
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Mr Davis 97 said:

Homework Statement


Solve: ##\displaystyle \lim_{x\rightarrow 2} \frac{\left | x^2 + 3x + 2 \right |}{x^2 - 4}##.

Homework Equations

The Attempt at a Solution


I am trying to solve the following limit: ##\displaystyle \lim_{x\rightarrow 2} \frac{\left | x^2 + 3x + 2 \right |}{x^2 - 4} = \lim_{x\rightarrow 2} \frac{\left | (x + 1)(x + 2) \right |}{(x - 2)(x + 2)}##. However, I am not sure how to proceed. Normally, I would cancel out the factors, but I am not sure what do with the absolute value in the numerator.
Since x is near 2, |x + 2| will be near 4; i.e., positive.
 
Mark44 said:
Since x is near 2, |x + 2| will be near 4; i.e., positive.
So I can just proceed as if there were no absolute value?
 
Mr Davis 97 said:

Homework Statement


Solve: ##\displaystyle \lim_{x\rightarrow 2} \frac{\left | x^2 + 3x + 2 \right |}{x^2 - 4}##.

Homework Equations

The Attempt at a Solution


I am trying to solve the following limit: ##\displaystyle \lim_{x\rightarrow 2} \frac{\left | x^2 + 3x + 2 \right |}{x^2 - 4} = \lim_{x\rightarrow 2} \frac{\left | (x + 1)(x + 2) \right |}{(x - 2)(x + 2)}##. However, I am not sure how to proceed. Normally, I would cancel out the factors, but I am not sure what do with the absolute value in the numerator.

Are you sure it's not supposed to be ##x^2 - 3x + 2## in the numerator?
 
PeroK said:
Are you sure it's not supposed to be ##x^2 - 3x + 2## in the numerator?
Pretty sure. Why?
 
Mr Davis 97 said:
Pretty sure. Why?

See whether you can work it out!
 
PeroK said:
See whether you can work it out!
Well I got DNE! Isn't that still a valid response?
 
Mr Davis 97 said:
Well I got DNE! Isn't that still a valid response?

For the original limit, assuming DNE means "does not exist", that's correct.

##\displaystyle \lim_{x\rightarrow 2} \frac{\left | x^2 - 3x + 2 \right |}{x^2 - 4}##.

Is a better problem, because it's of the form ##\frac{0}{0}## which the original problem wasn't.
 
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The point is that as x goes to 2, the denominator goes to 0 but the numerator does NOT. For x arbitrarily close to 0, the numerator close to 18, denominator close to 0, the fraction will be huge. If the limit were as x goes to -2, then it would be of the form "0/0".
 

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