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$$\lim_{x\rightarrow\infty}\sin^2(x^2)$$ I assume the limit doesn't exist but how do I show it?

Do I use sequences somehow?

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    $\begingroup$ Look at the function at $x=\sqrt{2n\pi}$ and $\sqrt{(2n+1)\pi/2}$. $\endgroup$ – André Nicolas May 22 '15 at 21:17
  • $\begingroup$ the limit dont exist. $\endgroup$ – abel May 22 '15 at 21:26
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As $x$ marches on toward infinity, it is reasonable to guess that the function $\sin^2(x^2)$ visits both 0 and 1 infinitely often. Can you construct an infinite sequence of values such that $\sin^2(x^2)$ evaluates to 0 for each term in the sequence? What about a similar sequence for which it evaluates to 1?

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    $\begingroup$ if i take a_n = root(2npi) and b_n = root((2n+1)pi/2), then both of these sequences tend to infinity as n tends to infinity, but f(a_n) tends to 0 as n tends to infinity, and f(b_n) tends to 1 as n tends to infinity, so because they are not equal, the limit does not exist. Does this work? $\endgroup$ – bws May 22 '15 at 21:34
  • $\begingroup$ @bws It sounds great, but I would add that $f(a_n)$ does not just tend toward 0, it is equal to 0 for all $n$. Similarly for $f(b_n)$. $\endgroup$ – Austin Mohr May 22 '15 at 21:37
  • $\begingroup$ oh, yes, my bad, thank you :) $\endgroup$ – bws May 22 '15 at 21:40
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If a function $f$ admits a limit, say $L$, as $x$ tends to $+\infty$ then for any sequence $(x_n)$ which tends to $\infty$, the sequence $(f(x_n))$ converges to $L$ too. However, if you take the two sequences $(x_n)$ and $(y_n)$ given by André Nicolas above, you will find that the two sequences $(f(x_n))$ and $(f(y_n))$ do not have the same limit. It follows that the limit do not exists, our function $f$ do not have a limit as tends to $+\infty$.

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One can even show the set of values of $\sin^2 n^2$ ($n$ is an integer) is dense in the interval $[0,1]$ (not trivial!).

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  • $\begingroup$ Actually more or less trivial, by continuity of the sine function. $\endgroup$ – Did May 22 '15 at 22:39
  • $\begingroup$ @Did: Actually, I was thinking of a nearby problem: when $x$ is a integer. $\endgroup$ – Bernard May 22 '15 at 22:44
  • $\begingroup$ Actually you might want to modify your post. $\endgroup$ – Did May 22 '15 at 22:44
  • $\begingroup$ @Did: done (pun intended :o)). $\endgroup$ – Bernard May 22 '15 at 22:47
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Note that the function $\sin^2 x$ never approaches a value as $x\to\infty$, it oscillates.

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  • $\begingroup$ A real function taking on the value $\;-1\;$ is highly unlikely. $\endgroup$ – Timbuc May 22 '15 at 22:08
  • $\begingroup$ @TravisJ Why the strange edit? $\endgroup$ – Did May 22 '15 at 22:40
  • $\begingroup$ @Did, the $-1$ was my mistake (as I was thinking $\sin(x)$ not $\sin^{2}(x)$. I removed the "never approaches a value" since that is false as $\lim_{x\to a}\sin^{2}(x)=\sin^{2}(a)$ for any $a\in \mathbb{R}$. $\endgroup$ – TravisJ May 22 '15 at 22:42
  • $\begingroup$ @Did, oscillates seems fine (as long as the $a$ is $\infty$)... as $x\to \infty$ the value of $\sin^{2}(x)$ oscillates between $0$ and $1$. Unless you feel like that asserts that it only takes on the values $0$ and $1$? $\endgroup$ – TravisJ May 22 '15 at 22:45

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