# Proving that the limit of a sequence lies within the range of the sequence. [duplicate]

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.

• At best, you can show $a\le x\le b$ Oct 31 '18 at 7:05

It is not true in general for example

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

therefore the correct statement is

$$a

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.