Prove that if $\lim_{n\to \infty} a_n=l$ then $\lim \sup a_n=\lim \inf a_n=l$

Question

Let $(a_n)_{n\in\mathbb{N}}\subset\mathbb{R}$, a bounded sequence. For each $n\in\mathbb{N}$, we have $A_n=\{a_k:k\ge n\}$. Let $\lambda_n=\sup A_n$ and $\beta_n=\inf A_n$.So we have $(\lambda_n)$ and $(\beta_n)$ a decresing and non-decreasing sequences, then the limits of those sequences are $\lim \sup$ and $\lim \inf$. Prove that if $\lim_{n\to \infty} a_n=l$ then $\lim \lambda_n=\lim \beta_n=l$. Any Hint? I'm lost.

Answer 1

Hint: If a sequence $\{a_n\}_n$ is convergent and it converges to $a$ every subsequence is convergent and it have limit equals to $a$.

Since $\limsup_na_n$ and $\liminf_na_n$ is the biggest and the smallest aderence value of $a_n$ you can conclude your result.

Answer 2

Depending on what you already know of $\limsup$ and $\liminf$, the question can be more or less difficult.

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For example, you may already know that $\limsup a_n$ is the largest limit point of $a_n$ and $\liminf$ is the smallest. In that case, the task is almost trivial.

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You can also go from the fact that $S=\limsup a_n$ is characterised by the property:

There are infinitely many elements of $a_n$ that are smaller than or equal to $S$, and for every $\epsilon>0$, there are only finitely many elements of $a_n$, greater than $S+\epsilon$.

In the case of a convergent sequence, it is simple to prove that the limit has the described property.