Simplifying lim_(n->oo) (root[n](2)-1)/(root[n](8)-1) I was attempting to go through a worked solution from my lecturer but I can't really understand this step: lim_(n->oo)(root[n](2)-1)/(root[n](8)-1)=lim_(n->oo}(1)/(root[n](4)+root[n](2)+1) What was actually even done here? And why? Any help is appreciated.

Jenny Stafford

Jenny Stafford

Open question

2022-08-19

Simplifying lim n 2 n 1 8 n 1
I was attempting to go through a worked solution from my lecturer but I can't really understand this step:
lim n 2 n 1 8 n 1 = lim n 1 4 n + 2 n + 1
What was actually even done here? And why?
Any help is appreciated.

Answer & Explanation

Kasen Schroeder

Kasen Schroeder

Beginner2022-08-20Added 21 answers

Hint. Note that
x 3 1 = ( x 1 ) ( x 2 + x + 1 )
which implies
2 3 n 1 = ( 2 n 1 ) ( 2 2 n + 2 n + 1 ) .
Damon Campos

Damon Campos

Beginner2022-08-21Added 1 answers

Hint:
2 n 1 8 n 1 = 2 n 1 8 n 1 4 n + 2 n + 1 4 n + 2 n + 1 = 8 n 1 ( 8 n 1 ) ( 4 n + 2 n + 1 )
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