How to factor y^1/3 - 1 and y^1/5 - 1 in a limit

[joshiemen]

Homework Statement

Lim x>1 (y^1/3-1)/((y^1/5-1)

Homework Equations

Difference of powers a^n-b^n=(a-b)*(a^n-1*b^0+a^n-2*b^1...+b^n-1)

The Attempt at a Solution

Scanned and attached the solution...

Can anyone Pls explain the steps in the solution, especially, how you factor out such an ugly function.

 

[Mentallic]

$$y-1\equiv (y^{1/3})^3-1\equiv (y^{1/n})^n-1$$, n being a natural number.

Does this clear things up?

 

[icystrike]

Homework Statement

Lim x>1 (y^1/3-1)/((y^1/5-1)

Homework Equations

Difference of powers a^n-b^n=(a-b)*(a^n-1*b^0+a^n-2*b^1...+b^n-1)

The Attempt at a Solution

Scanned and attached the solution...

Can anyone Pls explain the steps in the solution, especially, how you factor out such an ugly function.

you can actually substitute y=x^15 , as y>1 , x>1
hence obtaining x>1 (x^5-1)/(x^3-1)
try to divide x-1 from top and bottom

There is this identity:
x^n -1 =(x-1)[x^(n-1) + x^(n-2) + ... + 1 ]

Continue from here =D

 

[joshiemen]

you can actually substitute y=x^15 , as y>1 , x>1
hence obtaining x>1 (x^5-1)/(x^3-1)
try to divide x-1 from top and bottom

There is this identity:
x^n -1 =(x-1)[x^(n-1) + x^(n-2) + ... + 1 ]

Continue from here =D

:) THANK YOU!

Indeed, the joy of understanding is far greater and superior than the earthly pleasures one derives through out his/her life.

 

[joshiemen]

$$y-1\equiv (y^{1/3})^3-1\equiv (y^{1/n})^n-1$$, n being a natural number.

Does this clear things up?

Could you elaborate on this further pls? not so clear, i have been staring at it and thinking, have not penetrated the wall of ignorance yet.

 

[Mentallic]

Could you elaborate on this further pls? not so clear, i have been staring at it and thinking, have not penetrated the wall of ignorance yet.

It's essentially the same as icystrike's suggestion.

For $$y-1=(y^{1/3})^3-1$$ if you substitute $$x=y^{1/3}$$ or equivalently, $$x^3=y$$ then we have $$x^3-1=(x)^3-1$$

So if you have $$y-1$$ to factorize this using a difference of two cubes, you substitute $$x^3=y$$ or if you can see what is happening without substitution, you let $$y-1=(y^{1/3})^3-1^3=(y^{1/3}-1)(y^{2/3}+y^{1/3}+1)$$