# Sandwich theorem for shifted sequences proof verification

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 $$(y_{n+N}-l)\rightarrow 0\iff (y_{n+N})\rightarrow l$$.

$$\square$$

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 $$\lim_{n\to\infty} y_n=l$$.