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Factoring Involving Radicals

  • Thread starter Mach
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  • #1
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Homework Statement


This question is concerning a limit question. I have no problems finding limits but i need to be able to factor this equation. I need to be able to some how get rid of the (x-1)

(x^(1/6)-1)/(x-1)

The Attempt at a Solution


I ust keep ending up in a loop geting a bigger and bigger radical.
An out of curiosity can 0 be divided by 0?
 

Answers and Replies

  • #2
Gib Z
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Nope. Give us the actual limit, theres an easier way.
 
  • #3
HallsofIvy
Science Advisor
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No, you cannot divide by 0- not even 0! You should know, from a more general formula, that [itex]y^6- 1= (y- 1)(y^5+ y^4+ y^3+ y^2+ y+ 1)[/itex]. What happens if y= x^{1/6}?
 
  • #4
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No, you cannot divide by 0- not even 0! You should know, from a more general formula, that [itex]y^6- 1= (y- 1)(y^5+ y^4+ y^3+ y^2+ y+ 1)[/itex]. What happens if y= x^{1/6}?
Then all thats left is to create a new limit as y approaches something. In this case its still 1 but you would need to take the 6th root of the original limit point

For example, if you wanted to do something like the limit as x approaches 8 of:


[itex]\displaystyle{\frac{-x^2+ 4x^\frac{4}{3}+ x^\frac{7}{8}+ x^\frac{5}{6}- 4x^\frac{1}{2}- 4x^\frac{1}{6}}{x^\frac{2}{3}- 4}}[/itex]

You would have to again use y=x^(1/6), but take the new limit as y approaches 8^(1/6) or radical 2 of:
[tex]\displaystyle{\frac{-y^12+ 4y^8+ y^7+ y^5- 4y^3- 4y}{y^4- 4}}[/tex]
 
Last edited:

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