One-sided limits proof I need to prove this using sequences. How can I do that?  
If $\lim\limits{x\to a^-}f(x)=L$ and $\lim\limits_{x\to a^+}f(x)=L$, then $\lim\limits_{x\to a}f(x)=L$.
Thanks! 
 A: You need to know the definition of the limit at a real number. Assume that both $\lim_{x\to a^-}f(x)$ and $\lim_{x\to a^+}f(x)$ exist and eqauls to $L$. Then, given any $\epsilon>0$, there is a $\delta$ such that $|f(x)-L|<\epsilon$ for all $x$ such that $0<x-a<\delta$ or $0<a-x<\delta$( In fact for all $x$ such that $0<|x-a|<\delta$). But then $\lim_{x\to a}f(x)$ exists and it is equal to $L$.
A: $\lim_{x\longrightarrow a^{-}}f(x)=L$ means $\forall\{a_{n}\}_{n=1}^{\infty}\;s.t.\;\forall n\quad a_{n}\leq a\quad \{f(a_{n})\}_{n=1}^{\infty}\longrightarrow L$  . for the other limit we have same, that means for every sequence that all its members are greater than $a$ the sequence of their values by $f(x)$ converges to $L$. Now for proving the requested limit by sequences we should show if $\{a_{n}\}_{n=1}^{\infty}$ be any sequence that its elements are selected from $\mathbb{R}-\{a\}$ and is converges to $a$, the sequence of $\{f(a_{n})\}_{n=1}^{\infty}$ converges to $L$. But pay attention that the set $\{n\in\mathbb{N}|a_{n}< a\}$ and $\{n\in\mathbb{N}|a_{n}> a\}$ can not be finite at the same time, so one of them is infinite, think the first one is infinite, so the subsequence $\{b_{n}\}_{n=1}^{\infty}$ of $\{a_{n}\}_{n=1}^{\infty}$ that its elements are less than $a$ is exists, by the assumption of question $\{f(b_{n})\}_{n=1}^{\infty}$ is convergent to $L$. If the set $\{n\in\mathbb{N}|a_{n}> a\}$ be finite so $\{f(a_{n})\}_{n=1}^{\infty}$ except finitely many first elements is $\{f(b_{n})\}_{n=1}^{\infty}$ and it implies $\{f(a_{n})\}_{n=1}^{\infty}\longrightarrow L$. If not we have the other subsequence $\{c_{n}\}_{n=1}^{\infty}$ that all its elements are greater than $a$ but again by the question assumption we have $\{f(c_{n})\}_{n=1}^{\infty}\longrightarrow L$. now pay attention that $\{n\in\mathbb{N}|a_{n}< a\}\cup\{n\in\mathbb{N}|a_{n}> a\}=\mathbb{R}$. Since $\{f(b_{n})\}_{n=1}^{\infty}\longrightarrow L$ we have $\forall \varepsilon>0 \exists N_{1}\in\mathbb{N}\;s.t.\; \forall n\geq N_{1}\quad |f(b_{n})-L|<\varepsilon$ and as $\{f(c_{n})\}_{n=1}^{\infty}\longrightarrow L$ we have $\forall \varepsilon>0 \exists N_{2}\in\mathbb{N}\;s.t.\; \forall n\geq N_{2}\quad |f(c_{n})-L|<\varepsilon$. The first one makes that $\forall \varepsilon>0 \exists N_{1}\in\mathbb{N}\;s.t.\; \forall n\geq N_{1}, a_{n}<a\quad |f(a_{n})-L|<\varepsilon$ and the second one makes that $\forall \varepsilon>0 \exists N_{2}\in\mathbb{N}\;s.t.\; \forall n\geq N_{2}, a_{n}>a\quad |f(a_{n})-L|<\varepsilon$. By supposing $N:=max\{N_{1},N_{2}\}$ we have $\forall \varepsilon>0 \exists N\in\mathbb{N}\;s.t.\; \forall n\geq N\quad |f(a_{n})-L|<\varepsilon$ that is $\{f(a_{n})\}_{n=1}^{\infty}\longrightarrow L$ that complete the solution.
