prove that $\lim a_n = L$ if and only if $\limsup a_n = \liminf a_n = L$

Question

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Answer 1

Suppose $a_n \to L$, we show that among the real numbers $b$'s such that for every $\epsilon > 0$, there exists a natural number $N \ge 1$ such that $a_n < b+ \epsilon$ if $n > N$ , $L$ is the smallest. Suppose $K < L$ is a real number with the same property. Thus there exists a natural number $N_0$ such that: $L - a_n < \dfrac{L-K}{2} = \epsilon/2$, and $a_n - K < \dfrac{L-K}{2} = \epsilon/2$ if $n \ge N_0$. Thus: $L-K = (L-a_n)+(a_n-K)< \epsilon/2+\epsilon/2 = \epsilon = L-K$, contradiction. Thus $K = L$, and $L$ is the smallest number with the above property and is the lim sup of $a_n$'s. By the same argument, you can show it ( $L$ ) is the lim inf of the $a_n$'s and you have $L =$ lim sup $=$ lim inf. The other direction is easier and I am sure you can do it.