# Limit Superior and Limit Inferior

• Nov 11th 2011, 03:48 AM
Alexrey
Limit Superior and Limit Inferior
I was just wondering if my method for finding lim sup and lim inf for sequences is correct. It's different to how we do it in class, but I asked my lecturer about my way and he said that it looks fine but he'd think about it and get back to me. Essentially what I do is first take the sequence to infinity and then find its sup and inf at infinity, instead of finding the sup/inf first and then taking this to infinity. Here is an example to clear up what I'm saying:

Let the sequence x_n = (-1)^n + n/(n+1). What I first do is find the limit as n tends to infinity for each term. In this case, the first term alternates between 1 and -1, and the second term tends to 1. Once this is done I find the supremum and infimum. Therefore,

lim sup x_n = 1 + 1 = 2 and lim inf x_n = -1 + 1 = 0.

Is my method fine, or could there be a case where it would not work?
• Nov 11th 2011, 04:43 AM
Opalg
Re: Limit Superior and Limit Inferior
Quote:

Originally Posted by Alexrey
I was just wondering if my method for finding lim sup and lim inf for sequences is correct. It's different to how we do it in class, but I asked my lecturer about my way and he said that it looks fine but he'd think about it and get back to me. Essentially what I do is first take the sequence to infinity and then find its sup and inf at infinity, instead of finding the sup/inf first and then taking this to infinity. Here is an example to clear up what I'm saying:

Let the sequence x_n = (-1)^n + n/(n+1). What I first do is find the limit as n tends to infinity for each term. In this case, the first term alternates between 1 and -1, and the second term tends to 1. Once this is done I find the supremum and infimum. Therefore,

lim sup x_n = 1 + 1 = 2 and lim inf x_n = -1 + 1 = 0.

Is my method fine, or could there be a case where it would not work?

That method works fine, provided that one of the two sequences actually has a limit as $\displaystyle n\to\infty.$ In the case $\displaystyle x_n = (-1)^n + n/(n+1)$, the second sequence n/(n+1) tends to the limit 1, and the method works. But (to take a rather silly example) suppose that $\displaystyle x_n = (-1)^n + (-1)^{n+1}.$ In that case, each of the two terms has supremum 1 and infimum –1, and your method would suggest that $\displaystyle \limsup x_n = 2$ and $\displaystyle \liminf x_n = -2.$ But when you combine the two terms you see that $\displaystyle x_n=0$ for every n, and so $\displaystyle \limsup x_n$ and $\displaystyle \liminf x_n$ are both 0.