Limit of x Approaching 2: Squeeze Theorem and Rationalization | Wolfram Alpha

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Homework Help Overview

The problem involves evaluating a limit as x approaches 2, with a focus on the Squeeze Theorem and rationalization techniques. The original poster expresses uncertainty about how to apply the Squeeze Theorem and questions whether the limit exists.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss rationalizing the denominator and the potential use of conjugate factors. There is confusion regarding the application of these techniques, particularly whether to use the conjugate of the numerator, the denominator, or both.

Discussion Status

Some participants have offered guidance on using conjugate factors to address the limit, while others have expressed skepticism about the applicability of the Squeeze Theorem in this context. Multiple interpretations of the problem-solving approach are being explored.

Contextual Notes

There is mention of a specific textbook, Stewart's book, which may provide context for the problem. Participants are also reflecting on the origins of the problem and its commonality in educational resources.

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Homework Statement


[PLAIN]http://www4b.wolframalpha.com/Calculate/MSP/MSP32119hfegi8c73g07i600000h48cbcha52cc1ce?MSPStoreType=image/gif&s=5&w=78&h=46
As x approaches 2.

Homework Equations





The Attempt at a Solution


I rationalize the denominator and that doesn't work as it gives me another undefined answer. I just finished learning the Squeeze Theorem, so I'm guessing that I may have to use that, although I don't have a clue on how to apply it here. That or the limit simply doesn't exist. Am I way off? Where would the pros start with this one?
 
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Ah, I remember this one: is this out of Stewart's book? (If not, it's probably originally from the Russian collection "everyone" steals from...)

Did you try using a "conjugate factor" on the denominator, when you say you tried "rationalizing" it? Doesn't quite do the job, but it's a start. The "trick" is to also use a conjugate factor for the numerator: you want to get rid of the troublesome differences that go to zero. If you use the two conjugate factors, you'll find the troubles all clear up.

And I can tell you that the Squeeze Theorem is of no help here...
 
What dynamicsolo said works. The trick is rationalizing both numerator and denominator.
 
So do you mean to multiply the fraction by the conjugate of the denominator for both the numeratory and denominator, or is it to multiply the fraction by the conjugate of the numerator over the denominator?
 
NewtonianAlch said:
So do you mean to multiply the fraction by the conjugate of the denominator for both the numerator and denominator, or is it to multiply the fraction by the conjugate of the numerator over the denominator?

Neither.

multiply the fraction by the conjugate of the denominator for both the numerator and denominator

AND

multiply the fraction by the conjugate of the numerator for both the numerator and denominator
 
dynamicsolo said:
Ah, I remember this one: is this out of Stewart's book? (If not, it's probably originally from the Russian collection "everyone" steals from...)

Did you try using a "conjugate factor" on the denominator, when you say you tried "rationalizing" it? Doesn't quite do the job, but it's a start. The "trick" is to also use a conjugate factor for the numerator: you want to get rid of the troublesome differences that go to zero. If you use the two conjugate factors, you'll find the troubles all clear up.

And I can tell you that the Squeeze Theorem is of no help here...

It is from Stewart! Hah, good call. Thanks for the tip Dynamicsolo, I'll give it a go for sure.
 

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