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

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The limit of (x^2n - 1)/(x^2n + 1) as x approaches infinity is determined to be 1, as the dominant terms in the numerator and denominator cancel each other out. Substituting x with 1/h, where h approaches zero, also leads to the conclusion that the limit is 1. Discussions around the limit when n approaches infinity suggest it remains 1 for |x| > 1, 0 for |x| = 1, and -1 for |x| < 1, with no undefined values. The conversation highlights a common misunderstanding between limits approaching x versus n. Overall, the primary focus remains on the limit as x approaches infinity, confirming it equals 1.
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find lim (x^2n - 1)/(x^2n + 1)
x->infinity
-> means tends to
 
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lizzie said:
find lim (x^2n - 1)/(x^2n + 1)
x->infinity
-> means tends to

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


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.
 


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

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.
 


what will happen if n-> infinity
 


You asked that originally and you have already been given 4 answers.
 


DeaconJohn said:
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.

HallsofIvy said:
You asked that originally and you have already been given 4 answers.

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.
 
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Gib Z said:
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.

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
 
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