0

I have attempted a proof of the squeeze theorem which later turned up as a question in the book where the question was to find the flaw in this very proof.

If $x_n$ is convergent and $s_n \le x_n \le t_n$, and $s_n$ and $t_n$ converge to $L$, then given the order property of limits,

$\lim_{n\to \infty} s_n \le \lim_{n\to \infty} x_n \le \lim_{n\to \infty} t_n$

Thus $L \le \lim_{n\to \infty} x_n \le L$, and this can only be true if $\lim_{n\to \infty} x_n = L$

Where am I going wrong?

Aniruddh
  • 333
  • 4
    You need to show existence. As it stands, your argument could probably be fleshed out to show that ${x_n}$ can't approach any limit other than $L$, but that would still leave the possibility that it might not approach a limit at all. – lulu Jun 20 '22 at 16:30
  • 2
    @lulu "If $x_n$ is convergent" is at the start of the statement. – Arthur Jun 20 '22 at 16:33
  • 3
    @Arthur Ah, true. Missed that. Though the point of the Squeeze Theorem is that the assumptions imply the existence of the limit. – lulu Jun 20 '22 at 16:35
  • 1
    Which book is that? (I mean, what exactly does the question in the book look like? Is there some detail that you may have omitted from the above post?) –  Jun 20 '22 at 16:43
  • @StinkingBishop Elementary Real Analysis by Thomson and Bruncker – Aniruddh Jun 20 '22 at 16:45
  • 5
    Just looked it up. The Squeeze Theorem (Th. 2.20) does not assume up-front that $x_n$ is convergent - this is the conclusion of the theorem. The “careless student” (Ex. 2.8.3) does not assume that either, but somehow “passes the limit” through the inequality - that is the flaw. You do assume $x_n$ to be convergent, so your proof is correct, but what you have proven is not Squeeze Theorem but a weaker statement. Refer to some pdf of the book I found online. So @lulu ‘s guess was pretty much spot-on. –  Jun 20 '22 at 16:57

0 Answers0