# Find lim (x^2n - 1)/(x^2n + 1) x->infinity->

1. Jun 27, 2008

### lizzie

find lim (x^2n - 1)/(x^2n + 1)
x->infinity
-> means tends to

2. Jun 27, 2008

### aniketp

Re: Limits

Have you tried substituting x=1/h so as x->infinity, h->zero.
Using this I got the answer as 1

3. Jun 27, 2008

### CRGreathouse

Re: Limits

This is just (B - 1)/(B + 1) for some B that gets really big (goes to infinity with x). For all problems of this form you can just divide out the B and you'll see that the little stuff like +1 and -1 drop out.

4. Jun 27, 2008

### DeaconJohn

Re: Limits

Off the top of my head, my best guess is ...

For real numbers, the limit is equal to 1 if x > 1, -1/2 if x = 1, -1 if 0<=x<1, and is undefined if x<0.

For complex numbers with non-zero imaginary part, the limit is equal to -1 if |x| < 1 and is undefined if |x| >= 1.

5. Jun 28, 2008

### lizzie

Re: Limits

what will happen if n-> infinity

6. Jun 28, 2008

### HallsofIvy

Re: Limits

7. Jun 28, 2008

### Gib Z

Re: Limits

The original question asked for x--> infinity, not n.

As for the original question- Try adding and then subtracting 2 off the numerator.

For when n --> infinity, DJ had an attempt but needs some corrections: the limit is equal to one if |x| > 1, 0 if |x|=1, -1 if |x| < 1 and not undefined for any values. Note that we have an even function, so none of the "undefined if x<0" stuff.

Last edited: Jun 28, 2008
8. Jun 28, 2008

### DeaconJohn

Re: Limits

Gib Z, You are absolutely correct. My answer is in the context of n --> infinity. Funny thing is that I thought I was addressing the question of when x --> infinity. A credit to your understanding to realize that my calculations assumed n --> infinity. Unfortunately, I don't have anything to say about the case when x --> infinity, and the time that I've alloted to spend on this interesting problem of yours has expired.

I think I'll probably be concentrating on the number theory board in the future. That is the area of math that interests me most these days. It is also the area I know least about.

DJ