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Does the $\lim_{x \rightarrow 0} \sin^2(\frac{1}{x})$ exist? My understanding is that $$\lim_{x \rightarrow 0} \sin^2\left(\frac{1}{x}\right) = \lim_{x \rightarrow 0} \sin\left(\frac{1}{x}\right) \cdot \lim_{x \rightarrow 0} \sin \left(\frac{1}{x}\right) = DNE.$$ Is there any better way to view or solve it?

Robert Z
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Jaynot
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    Technically it's not enough, because the product of two functions without limits may still have a limit. The fact that the two are equal helps, but not entirely. The fact that they are equal and continuous might be enough. – Arthur Sep 19 '17 at 20:01
  • There certainly is a better way to solve it, since this is not a solution. – zhw. Sep 20 '17 at 23:13
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    It is NOT generally true that if $L=\lim_{x\to a}f(x)g(x)$ then $L=(\lim_{x\to a}f(x))\cdot (\lim_{x\to a}g(x)).$.... It is true IF $\lim_{x\to a}f(x)$ and $\lim_{x\to a}g(x)$ exist in $\mathbb R.$ .... For example let $f(x)=0$ if $x\in \mathbb Q$ and $f(x)=1$ if $x\in \mathbb R$ \ $\mathbb Q.$ Let $g(x)=1-x.$ Then $f(x)g(x)=0$ for all $x$ so $0=\lim_{x\to 0}f(x)g(x).$ But $\lim_{x\to 0}f(x)$ and $\lim_{x\to 0}g(x)$ don't exist. – DanielWainfleet Sep 21 '17 at 00:04

2 Answers2

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Hint. Consider the limit along these two sequences: for $n\geq 1$ $$x_n=\frac{1}{\pi n},\quad y_n=\frac{1}{\pi (n+1/2)}.$$ We have that $x_n\to 0$ and $y_n\to 0$. What are the limits: $$\lim_{n\to \infty} \sin^2(1/x_n)=?\quad\mbox{and}\quad \lim_{n\to \infty} \sin^2(1/y_n)=?$$

Robert Z
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  • $ \lim \sin^2( \frac{1}{x_n}) = 0$ as n goes to infinity @Robert and the second limit is 1 – Jaynot Sep 19 '17 at 20:13
  • @Jaynot Well done! Since these two limits are different, it follows that $\lim_{x \rightarrow 0} \sin^2(\frac{1}{x})$ does not exist. – Robert Z Sep 19 '17 at 20:30
  • Thank you. Here is the actual problem $\lim_{x \rightarrow 0} \frac{1}{x}\int_{0}^{x}\sin^2(\frac{1}{u})du$ that I was trying to solve. Does that suffices to say that the limit does not exceed? I would appreciate a little hint. Since the limit of $ \sin^2( \frac{1}{x}$ does not exist, I am solving some trouble figuring this out. – Jaynot Sep 19 '17 at 20:34
  • @Jaynot No, the limit of with the integral is a different thing. If after applying Hopital the limit does not exist then this does NOT imply that the original limit does not exist too. The limit of with the integral should be $1/2$. – Robert Z Sep 19 '17 at 20:49
  • @Jaynot Note that $\sin(t)^2=(1-\cos(2t))/2$. – Robert Z Sep 19 '17 at 21:11
  • Got it. Thanks. – Jaynot Sep 19 '17 at 21:13
  • @Jaynot: some more work is needed (after hint of Robert Z) to solve the limit of the integral. See this question https://math.stackexchange.com/q/1551332/72031 – Paramanand Singh Sep 20 '17 at 01:26
  • @ParamanandSingh Please see the new answer that I am adding below based on the new question. – Jaynot Sep 20 '17 at 16:18
  • @RobertZ you can also please check. – Jaynot Sep 20 '17 at 16:20
  • @Jaynot: I can see your deleted solution which is correct but needs to have more steps. The limit of $\frac{1}{x}\int_{0}^{x}\cos(2/u),du$ as $x\to 0$ is $0$, but this fact is non-trivial and needs to be shown. The linked question in my previous comment deals with this problem. So you need to put some argument like that. It would be better if you ask this as a new question. – Paramanand Singh Sep 21 '17 at 04:53
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Let the limit be $L$.

Then the equation $$\sin^2\frac1x=\left(L+\frac12\right)\bmod1$$ has infinitely many solutions in the neighborhood of $0$, such as $$x_k:=\dfrac1{\arcsin\sqrt{(L+\frac12)\bmod1}+2k\pi},$$ and

$$\left|\sin^2\frac1{x_k}-L\right|\ge\frac12.$$