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If $X=(x_n)$ and $Y=(y_n)$ are sequences of real numbers which both converge to $c$ and if $Z=(z_n)$ is a sequence such that $x_n\le z_n\le y_n$ for $n\in \mathbb{N}$, then $Z$ also converges to $c$.

What I know is that $\text{lim}_{n\to\infty}(x_n)=c$ and that $\text{lim}_{n\to\infty}(y_n)=c$. Thus I know by the definition of a limit, that for any $\epsilon>0$ there exists $N$, $\forall$ $n>N$ such that $$\left|{x_n}-c\right|<\epsilon$$ $$\left|{y_n}-c\right|<\epsilon.$$ How can I continue? I know that any subsequence of a convergent sequence must converge. I also know, $x_n\le z_n\le y_n \implies lim(x_n)\le lim(z_n)\le lim(y_n)$, but I do not know how to incorporate all of this together in completing my proof?

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for any $\varepsilon>0$, there exists $N$ s.t. if $n>N$ then $|x_n-c|<\varepsilon$, $|y_n-c|<\varepsilon$ for all $n$. So if $n>N$, then $$c-\varepsilon < x_n<c+\varepsilon$$ and $$c-\varepsilon <y_n<c+\varepsilon$$ for all $n$. So $$c-\varepsilon < x_n \le z_n \le y_n < c+\varepsilon$$ and we get $$|z_n-c|<\varepsilon$$ for all $n>N$.

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