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Suppose $x_n$ is a bounded sequence and $\limsup x_n = \liminf x_n = c$. Prove $x_n \rightarrow c, n \rightarrow \infty$

Not sure if there is a better way to show this, or if my way is even correct, just looking for some guidance or tips.

Since $\limsup x_n = c$, then given $\epsilon > 0$, there exists a $N_1$ such that for all $n \geq N_1$ we have:

$x_n < c + \epsilon$.

Likewise, since $\liminf x_n = c$, then there exists some $N_2$ such that for all $n \geq N_2$ we have:

$c - \epsilon < x_n$

Let $N = \max\{N_1, N_2\}$, then for all $n \geq N$, we get:

$c - \epsilon < x_n < c + \epsilon$

$|x_n - c| < \epsilon$

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Your proof is correct. You might write down the definition of $\limsup$ before concluding that $x_nc-\epsilon$.

$$\limsup_{n\to\infty}x_n=c\iff\forall \epsilon>0\ \exists N_1\in\mathbb N\ \forall n\ge N_1, -\epsilon<\sup\{x_n,x_{n+1}\ldots\}-c<\epsilon$$

After this, you may safely conclude that $x_n

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    I'm sorry, not quite sure I understand what I should do? – 2017-01-29
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    I explained in more detail. – 2017-01-29
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    Oh I see, thanks! Quick question, isn't: $-\epsilon < \sup\{x_n, x_{n+1}, ...\} -c$ true for all n? – 2017-01-29
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    Right, because $\sup\{x_n,x_{n+1}\ldots\}$ decreases to $c$. – 2017-01-29