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

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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.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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