Relation limsup, liminf, lim

Consider a sequence of real numbers $\{X_n\}_{n \in \mathbb{N}}$ and suppose that I have shown that

$\forall \epsilon>0$, $\limsup_{n\rightarrow \infty} X_n-X\leq\epsilon$ and $\liminf_{n\rightarrow \infty} X_n-X\geq-\epsilon$, $X \in \mathbb{R}$

Questions:

1) Does this imply $\lim_{n \rightarrow \infty} X_n=X$?

2) Why yes or not?

• What have you tried so far? Do you know any properties of $\lim\inf$ and $\lim\sup$ or just the basic definitions? Nov 27, 2015 at 13:51

The relations $\limsup X_n-X\leq\varepsilon$ and $\liminf X_n-X\geq-\varepsilon$ imply $$\limsup X_n-\liminf X_n\leq2\varepsilon$$ for all $\varepsilon>0$; thus $X=\limsup X_n=\liminf X_n$.
• Because $\limsup X_n-X\leq\varepsilon$ for all $\varepsilon>0$ implies $\limsup X_n=X$ (it is the definition of limit). Similarly, $\limsup X_n=\liminf X_n$. Nov 27, 2015 at 14:16
• I still don't understand. You say that $\limsup_{n \rightarrow \infty} X_n-X\leq \epsilon$ for all $\epsilon>0$ imply that $\limsup_{n \rightarrow \infty}X_n =X$ using the definition of limits. However, using the definition of limits and defining $A:=\limsup_{n\rightarrow \infty} X_n$ I can say that $\forall \epsilon>0$ $\exists$ $\bar{n}$ such that $|sup_{k\geq n} X_k-A|<\epsilon$ $\forall n \geq \bar{n}$. How do you infer that $A=X$?
• Sorry, it was my mistake. $\limsup X_n-X\leq\varepsilon$ for all $\varepsilon>0$ not necessarily implies $\limsup X_n=X$. But both inequalities imply $$X-\varepsilon\leq\liminf X_n\leq\limsup X_n\leq X+\varepsilon.$$ Since $\varepsilon$ is arbitrary, you have $$X\leq\liminf X_n\leq\limsup X_n\leq X;$$ therefore $\liminf X_n=\limsup X_n=X$. Nov 27, 2015 at 18:26