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I'm trying to proof a sandwich theorem for shifted sequences.

Given $(x_n)\rightarrow l, (z_n)\rightarrow l$ and $x_n\le y_n \le z_n$ eventually, show that $(y_n)\rightarrow l$.

My attempt is as follows. I'll rewrite "$x_n\le y_n \le z_n$ eventually" as "$x_{n+N}\le y_{n+N} \le z_{n+N} \iff 0\le y_{n+N}-x_{n+N} \le z_{n+N}-x_{n+N}.$

Now since $|z_{n+N}-l|\lt\epsilon$ and $|x_{n+N}-l|\lt\epsilon$, $\forall n>N\in\mathbb{N}$ it's reasonable to say that $z_{n+N}-x_{n+N}=0$.

And so by the sandwich rule for a null sequences $(y_{n+N}-l)\rightarrow 0\iff (y_{n+N})\rightarrow l$.

$\square$

Any comments are appreciated.

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If by $x_n\leqslant y_n\leqslant z_n$ eventually you mean there exists $N$ such that $x_n\leqslant y_n\leqslant z_n$ for $n\geqslant N$, then yes, this is true. Let $\varepsilon>0$ and choose $N_x>N$ such that $|x_n-l|<\varepsilon$ for $n\geqslant N_x$, and $N_z>N$ such that $|z_n-l|<\varepsilon$ for $n\geqslant N_z$. Then for $n\geqslant\max\{N_x,N_z\}$ we have $$ y_n-l\leqslant z_n-l<\varepsilon $$ and $$ -(y_n-l)\leqslant -(x_n-l)<\varepsilon, $$ so that $|y_n-l|<\varepsilon$ and $\lim_{n\to\infty} y_n=l$.

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