# Limits splitting the fraction into two

#### StephenPrivitera

limn-->oo(x2n-1)/(x2n+1)
I can't figure this one out. I've tried everything. I tried splitting the fraction into two, applying a log to each side, factoring the top, dividing by x2n, and Lhopitals rule doesn't apply and wouldn't help if it did. Any ideas?

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#### StephenPrivitera

perhaps the only way to do this would be to consider various cases such as x>1, x=1, x<-1, x=-1, etc.
anyone agree?

I get for x>1
lim=1
for x=1, lim=0
if 0<x<1, lim=-1

What about x<0? What is -2^999999.5? Surely, it's not real. Can we say that lim DNE for x<0?
For -1<x<0 x^n would be very small, but wouldn't n have to be some integer?

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#### StephenPrivitera

I got stuck.

For example,
I'll omit the subscripts.

lim(x2n-1)/(x2n+1)=lim(1-1/x2n)/(1+1/x2n)
As n grows to infinity, we can say nothing about the limit, because it depends on what x is. If x is small then 1/x is large. If x is big, then 1/x is small.

edit: whoops edited wrong post, sorry.

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#### Hurkyl

Staff Emeritus
Gold Member
Grargh, you read and responded before I could delete my post.

Yes, breaking it up into cases is a good idea.

edit: n often implicitly means an integer, and it wouldn't surprise me if this problem assumed as such.

*sigh* Today isn't my best day.

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#### HallsofIvy

Homework Helper
I would interpret this as n being an integer so you don't have any problems with fractional powers of negative numbers.

Because x2n= (x2)n) it doesn't matter whether x is positive or negative so you might as well assume positive. In that case the crucial cases are: 0<= x<1, x= 1, x> 1.

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