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If $\lim x_n$ and $\lim y_n$ exists and if $x_n\leq w_n\leq y_n$ for each $n\in\mathbb{N}$ then $\lim w_n$ also exists.

Initially, I assumed that this is a weaker statement of the squeeze theorem for sequences and that it was true, but have learned that the statement is false. What about this statement makes it false?

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It is false because, for instance, if you take $x_n=-1$, $y_n=1$, and $w_n=(-1)^n$, you indeed have$$(\forall n\in\mathbb{N}):x_n\leqslant w_n\leqslant y_n,$$but the limit $\lim_{n\to\infty}w_n$ does not exist.

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  • $\begingroup$ Ah, I see now that why simply stating that the limit exists is not enough. Thank you! $\endgroup$ – Peetrius Jul 8 '18 at 17:22
  • $\begingroup$ @Peetrius It's true if you assume that $\lim x_n=\lim y_n$. It's also true if you assume that $\lim w_n$ exists... $\endgroup$ – David C. Ullrich Jul 8 '18 at 17:29

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