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**Abu-Khalil** Let $\displaystyle \epsilon>0$. Suppose $\displaystyle a_n\to A\in\mathbb{R}$, then $\displaystyle \exists N:\forall n\geq N,|a_n-A|<\epsilon$.

Let $\displaystyle r=\max_{1\leq n\leq N}|a_n|$ then choose $\displaystyle R=\max\left\{r+\epsilon,|A|+\epsilon,\right\}.$

The $\displaystyle r+\epsilon$ is to get a new $\displaystyle n$ and then, for a $\displaystyle n\geq N$ we have

$\displaystyle \epsilon>|a_n-A|\geq |a_n|-|A|\geq R+|A|+\epsilon-|A|=R+\epsilon.\Rightarrow 0>R\rightarrow\leftarrow$