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?


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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