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I vaguely recall from undergraduate analysis that there is a relationship that allows sequences to be used to show that certain limits don't exist. For example, I recently came across a limit problem, specifically the limit of $\sin(1/x)$ as $x$ approaches zero. The person who posted the solution used sequences to show that this limit does not exist. I was wondering if someone would be kind enough to refresh my memory and generalize this process along with any theorems that may be necessary or perhaps refer me to another internet resource that would do so, as I was foolish enough to sell my analysis text, and I don't believe a basic undergraduate calculus text would cover this topic.

Thanks

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    $\begingroup$ F(x) converges to as x goes to x0 iff for every sequence yn going x0, F(yn) converges to a unique limit $\endgroup$ – Lost1 Dec 27 '13 at 13:30
  • $\begingroup$ Use wikipedia for "limit of a function" and "continuity". I'm not sure that the wikipedia article on "singularity" is readable enough, but in general the exceptions to continuity are treated under this heading "singularities of real functions". The concrete example is discussed under "Topologist's sine curve". $\endgroup$ – Dr. Lutz Lehmann Dec 27 '13 at 15:22
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For any fixed $a\in[-1,1]$, the equation $f(x)=\sin(1/x)=a$ has infinitely many solutions on every interval $(0,\delta)$, from which you can construct a sequence $x_n\to 0$ such that $\lim_{n\to\infty} f(x_n)=a$. Since you can reach any $a\in[-1,1]$, the limit as a function does not exist. This applies to any oscillating singularity.

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  • $\begingroup$ @Did, care to explain? $\endgroup$ – Dr. Lutz Lehmann Dec 27 '13 at 14:09
  • $\begingroup$ Sure. Read the question. $\endgroup$ – Did Dec 27 '13 at 14:35
  • $\begingroup$ Then read again my answer and how it explains how to construct such sequences. $\endgroup$ – Dr. Lutz Lehmann Dec 27 '13 at 15:16
  • $\begingroup$ My first comment was deleted (I wonder why), so here it is again: "Off-topic." To belabor the obvious, the OP's query is not about the function $f:x\mapsto\sin(1/x)$ per se (whose case they seem to know) but about the more general setting explicitely described in the title (and in the text). Thus, Lost1's comment seems fully on-topic. $\endgroup$ – Did Dec 27 '13 at 20:51
  • $\begingroup$ The point that Dud skipped over is that the graph of sin(1/x) does not only have two points (as it is usually demonstrated) as accumulation points, but the whole segment [-1,1]. Which makes it different from a jump discontinuity. $\endgroup$ – Dr. Lutz Lehmann Dec 28 '13 at 0:47

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