I don't really understand how to find the infimum and supremum and the limsup and liminf of a sequence.
$x_n$=$(-1)^n+\frac{1}{n}$
$x_n$=$\frac{(-1)^n}{n}$
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Let's start with infimum and supremum. These are, respectively, the greatest lower and least upper bounds. That is, the infimum is the largest value that your sequence is never less than, and the supremum is the smallest value that your sequence is never greater than. That's pretty abstract, so let's look at your sequences to see what that means in concrete terms. For your first sequence we have $x_n = (-1)^n+1/n$. Looking at the first few terms we get: $$x_1 = 0, x_2 = 3/2, x_3 = -2/3, x_4 = 5/4, \ldots$$
This alternates, so it's worth looking at the negative terms (when $n$ is odd) and the positive terms (when $n$ is even) separately. For the odd terms, the first few are: $$x_1 = 0, x_3 = -2/3, x_5 = -4/5, x_7 = -6/7 \ldots$$ This looks like it's going to converge to $-1$, and you can check that it does for yourself. Since this is all the negative terms, the sequence can never get smaller than $-1$, so that's going to be our infimum: the sequence will never actually get to $-1$, but you can choose $n$ large enough, provided it's odd, that you can get as close as you like to $-1$.
If it actually reached $-1$? Well then it's still the infimum, but it's also then the minimum as well. As it is, this sequence has no minimum.
Back to the terms where $n$ is even; the first few are: $$x_2 = 3/2, x_4 = 5/4, x_6 = 7/6, x_8 = 9/8, \ldots$$ So it looks as though this is also converging towards $1$ (and you can check that it is), so our supremum must be... 3/2. This is also the maximum, but the sequence never gets any bigger, so it must be the supremum.
Lim-inf and lim-sup are now the limits of the sequence of the infima and suprema respectively. When we split the sequences up into the positive and negative parts that helps us here, because the lim inf only needs to look at the negative values (since they're all smaller than the positive ones!). So the lim-inf is $$\lim (\inf\{x_1\}, \inf\{x_1, x_3\}, \inf\{x_1, x_3, x_5\}, \ldots)$$ i.e. $$\lim (\inf\{0\}, \inf\{0, -2/3\}, \inf\{0, -2/3, -4/5\}, \ldots)$$ and you should be able to see that this is converging towards $-1$.
You should be able to handle the lim-sup case now.
The same techniques will work for the second sequence as well, and since Razieh has provided the answers for you, I'll leave this here.
postmortes
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So for the second one, is my inf=0 and sup=1/2 – ematth7 Sep 19 '15 at 16:20
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Yep, exactly. Sounds like you've got it! – postmortes Sep 19 '15 at 16:23
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Thank you for your help and explanations! – ematth7 Sep 19 '15 at 16:26
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@ematth7 no for second one theorem:Let $(x_n)$ be a bounded sequence in R. Then $(x_n)$ converges if and only if $\lim sup x_n = \lim inf x_n$. Here second sequence converges. – R.N Sep 19 '15 at 16:51
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@RaziehNoori - ematth7 asked about inf and sup in their comment, not lim-inf and lim-sup. – postmortes Sep 19 '15 at 17:56
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oh, you are right, my mistake. – R.N Sep 19 '15 at 17:59
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For first one $\limsup=1$ and $\liminf=-1$, and for second one $\limsup = \liminf= \lim x_n =0$.
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@ematth7 second one is obvious. for first one it is enough to write it for $n=1, 2, 3, ...$ to manage your mind – R.N Sep 19 '15 at 15:56
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@ematth7 see this http://math.stackexchange.com/questions/493526/can-some-clearly-explain-about-lim-sup-and-lim-inf – R.N Sep 19 '15 at 16:46