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If $a < x_{n} < b$ and $\lim_{n \rightarrow \infty}x_{n}=x$ then prove that $a \leq x \leq b$. I am taking two case where the sequence is monotonic and non-monotonic. If the sequence is increasing then it would converge to its supremum and if its decreasing then it will converge to its infimum, hence the limit being greater than $a$ and less than $b$. But I am not sure how to go on about the non-monotonic case.

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    $\begingroup$ At best, you can show $a\le x\le b$ $\endgroup$ Oct 31 '18 at 7:05
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It is not true in general for example

$$0<\frac1n<1 \quad \frac1n \to 0$$

therefore the correct statement is

$$a<x_n<b \quad x_n \to x \implies a\le x\le b$$

To prove we can simply assume $\epsilon$ such that $(x-\epsilon,x+\epsilon)\subseteq(a,b)$ that is

$$|x-\epsilon|\le \min\{|x-a|,|x-b|\}$$

and then apply the definition of limit.

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