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I have the following function which I'm considering on $[0,1)$ $$g(\theta) = \frac{1}{2\pi^2\theta^3}-\frac{\pi}{2}\cot(\pi\theta)\csc^2(\pi\theta).$$ According to a graph in mathematica it is continuous at $$\theta=0.$$I want to prove this rigorously by showing it has a finite limit. I've thought about using L'Hopital's rule but the expressions are too complicated. Any suggestions?

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  • $\begingroup$ Using the L'Hoital's rule shows that you are reluctant to use $\epsilon-\delta$ to prove the continuity. Right? $\endgroup$ – mrs Mar 24 '13 at 12:08
  • $\begingroup$ I can't really see an easy way to do that either $\endgroup$ – user67881 Mar 24 '13 at 12:11
  • $\begingroup$ Have you tried using Taylor expansion which is base on using derivatives? $\endgroup$ – mrs Mar 24 '13 at 12:14
  • $\begingroup$ That results in a similar problem but with the limits of the derivatives. $\endgroup$ – user67881 Mar 24 '13 at 12:25
  • $\begingroup$ Check this technique. $\endgroup$ – Mhenni Benghorbal Mar 24 '13 at 13:36
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Hint We have $$\cot(\pi t)=\frac{1}{\pi t}-\frac{1}{3}\pi t+O(t^3)$$ $$\csc^2(\pi t)=\frac{1}{\pi^2 t^2}+\frac{1}{3}+O(t^2)$$ so we find $$g(t)=_{t\to 0}O(t)$$ and thus $$\lim_{t\to 0}g(t)=0.$$

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    $\begingroup$ @Sami $\sin(x)\approx x$ around $0$, but $\sin (x) -x\not \approx 0$ around $0$. You can't just subtract. $\endgroup$ – Git Gud Mar 24 '13 at 12:36
  • $\begingroup$ @GitGud I didn't subtract approximations but equalities. You see? $\endgroup$ – user63181 Mar 24 '13 at 13:02
  • $\begingroup$ @SamiBenRomdhane Hm... ok. I guess I see that. I don't think that point is entirely clear, though. $\endgroup$ – Git Gud Mar 24 '13 at 13:11
  • $\begingroup$ You see? No, sorry but I do not. $\endgroup$ – Did Mar 24 '13 at 13:49
  • $\begingroup$ @Did in the example given by Git, it's true that we can't write $\sin x-x\sim 0$ even we have $\sin x\sim x$ because we deal with approximation but we have $\sin x=x+o(x)$ so we also have $\sin x-x =o(x)$ because here we deal with equality. $\endgroup$ – user63181 Mar 24 '13 at 13:59
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Note that $g(\theta)=\pi h(\pi\theta)/k(\pi\theta)$ with $k(x)=2x^3\sin^3(x)$ and $h(x)=\sin^3(x)-x^3\cos(x)$.

First, one term in the expansion of sine at zero is $\sin(x)=x+O(x^3)$, which yields $k(x)\sim2x^6$. Second, two terms in the expansion of cosine at zero are $\cos(x)=1-x^2/2+O(x^4)$. Third, two terms in the expansion of sine at zero are $\sin(x)=x-x^3/6+O(x^5)$, which yield $$ \sin^3(x)=(x-x^3/6+O(x^5))^3=x^3(1-x^2/6+O(x^4))^3=x^3(1-x^2/2+O(x^4)). $$ Thus, $h(x)=x^3\cdot O(x^4)=O(x^7)$, hence $h(x)/k(x)=O(x)$. This implies in particular that $g(\theta)\to0$ when $\theta\to0$.

Edit: And $g'(0)=\frac\pi{30}\ne0$ hence $g(\theta)=\Theta(\theta)$.

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    $\begingroup$ @Stephen I recommend you take a look at this answer. $\endgroup$ – Git Gud Mar 24 '13 at 12:49
  • $\begingroup$ @GitGud Isn't it essentially the same argument? $\endgroup$ – user67881 Mar 24 '13 at 12:55
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    $\begingroup$ @Stephen I don't think so. Note that this answer transform $g$ into a rational function first. I believe that's a crucial point. For instance, around $0$ you have $\sin(x)\sim x$, but $\frac{1}{x}-\frac{1}{\sin (x)}\neq \sim \frac{1}{x} - \frac{1}{x}=0$. Correct would be $\displaystyle \frac{1}{x}-\frac{1}{\sin (x)}=\frac{\sin (x)-x}{x\sin (x)}\sim \frac{-(1/6)x^3}{x^2}=-\frac{1}{6}x$ $\endgroup$ – Git Gud Mar 24 '13 at 13:04
  • $\begingroup$ @GitGud To clarify, the point I was making was that the argument takes Taylor series and combines them in an appropriate way as SamiBenRomdhane's answer did. But your point is still an important warning. $\endgroup$ – user67881 Mar 24 '13 at 13:25
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    $\begingroup$ @Stephen If I may enter the discussion... To tell you the truth I was rather surprised to see Sami's answer accepted. To my eyes, the line "so we find" is where everything is hidden. More precisely, I wonder how one can ask the question AND be able to convert "so we find" into a full solution. Or, putting it another way, I fail to see where the other solution "combines" anything at all "in an appropriate way". (Having said that, I should probably recall that votes and acceptances are entirely free on the site.) $\endgroup$ – Did Mar 24 '13 at 13:47
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Or; when $\theta$ tends to $0$, we know that $$\tan(\pi\theta)\sim\pi\theta,~~ \sin(\pi\theta)\sim\pi\theta$$ Hence your function tends to zero.

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    $\begingroup$ You don't have to apologize. I think your answer suffers from the same issues as Sami's. $\endgroup$ – Git Gud Mar 24 '13 at 20:56
  • $\begingroup$ @GitGud: Thanks for saying so, however, the points you and Did noted covered the answer more than the OP had expect. Prime points about the very small functions.:) $\endgroup$ – mrs Mar 24 '13 at 20:59
  • $\begingroup$ @GitGud Hi I've just looked back at this again and I was wondering if you could clarify one more point for me. Didn't Did make the mistake you were trying to point out in Sami's answer. Specifically Did states $2x^3\sin^3(x) \sim 2x^6$ and then later divides. $\endgroup$ – user67881 Mar 24 '13 at 23:43
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    $\begingroup$ @Stephen Nop, because multiplication offers no problems. Just addition (and subtraction) do. But you're better of asking Did directly anyway. He's really good at this. $\endgroup$ – Git Gud Mar 24 '13 at 23:47

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