How can factoring help solve limit problems involving radicals?

[Mach]

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?

 

[Gib Z]

Nope. Give us the actual limit, there's an easier way.

 

[HallsofIvy]

No, you cannot divide by 0- not even 0! You should know, from a more general formula, that $y^6- 1= (y- 1)(y^5+ y^4+ y^3+ y^2+ y+ 1)$. What happens if y= x^{1/6}?

 

[turdferguson]

No, you cannot divide by 0- not even 0! You should know, from a more general formula, that $y^6- 1= (y- 1)(y^5+ y^4+ y^3+ y^2+ y+ 1)$. What happens if y= x^{1/6}?

Then all that's 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: $\displaystyle{\frac{-x^2+ 4x^\frac{4}{3}+ x^\frac{7}{8}+ x^\frac{5}{6}- 4x^\frac{1}{2}- 4x^\frac{1}{6}}{x^\frac{2}{3}- 4}}$

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:
$$\displaystyle{\frac{-y^12+ 4y^8+ y^7+ y^5- 4y^3- 4y}{y^4- 4}}$$