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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?

  • Using the L'Hoital's rule shows that you are reluctant to use $\epsilon-\delta$ to prove the continuity. Right? – Mikasa Mar 24 '13 at 12:08
  • I can't really see an easy way to do that either –  Mar 24 '13 at 12:11
  • Have you tried using Taylor expansion which is base on using derivatives? – Mikasa Mar 24 '13 at 12:14
  • That results in a similar problem but with the limits of the derivatives. –  Mar 24 '13 at 12:25
  • Check this technique. – Mhenni Benghorbal Mar 24 '13 at 13:36
  • Stephen: Just for the record: to ask for explanations about answer A in a comment to answer B implying that answer A is wrong, as you did below, is in very bad taste. One thing is to debate about the merits of your acceptance strategy (which some find surprising) but to cast aspersions on other answers, unbeknowst to the answerers, is entirely different, and quite inadequate. – Did Mar 29 '13 at 07:50
  • @Did Point taken there. However, does "some finding surprising" relate to my acceptance strategy in general? –  Mar 29 '13 at 12:37
  • Sorry but I do not understand your last question. If, as I suspect, it is mostly rhetorical, I suggest to leave it aside. – Did Mar 29 '13 at 12:51
  • @Did I just don't feel it was very appropriate to raise the point about my acceptance strategy in that comment. –  Mar 29 '13 at 13:22
  • You are derailing the subject. As you are well aware, the point was that you "cast aspersions on other answers, unbeknowst to the answerers". (The acceptance strategy is quite peculiar as well, but is not the point.) Again, bad manners. – Did Mar 29 '13 at 13:39
  • @Did As I said I accepted your point about how I "cast aspersions on other answers, unbeknowst to the answerers" and in the future I will avoid this. However, I strongly disagree with your comment of "bad manners" and in general your comments throughout this post on the way I have chosen my answer. –  Mar 29 '13 at 13:44
  • Your function is actually undefined at $\theta=0$. You aren't seeking to show that $g$ is continuous at $0$. You are seeking to show that $\lim\limits_{\theta\to0^{+}}g(\theta)$ exists. – 2'5 9'2 Jan 04 '16 at 07:43

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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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    @Sami $\sin(x)\approx x$ around $0$, but $\sin (x) -x\not \approx 0$ around $0$. You can't just subtract. – Git Gud Mar 24 '13 at 12:36
  • @GitGud I didn't subtract approximations but equalities. You see? –  Mar 24 '13 at 13:02
  • @SamiBenRomdhane Hm... ok. I guess I see that. I don't think that point is entirely clear, though. – Git Gud Mar 24 '13 at 13:11
  • You see? No, sorry but I do not. – Did Mar 24 '13 at 13:49
  • @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. –  Mar 24 '13 at 13:59
  • All these are true. And your point is? – Did Mar 24 '13 at 14:01
  • @GitGud: Dear Gid, It seems that my attempt was not enough to handle this question. Cause it got -1. So, am I supposed to remove it? Thanks and sorry for asking. – Mikasa Mar 24 '13 at 20:48
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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 $g(\theta)=\Theta(\theta)$.

Did
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    @Stephen I recommend you take a look at this answer. – Git Gud Mar 24 '13 at 12:49
  • @GitGud Isn't it essentially the same argument? –  Mar 24 '13 at 12:55
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    @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 $\displaystyle \frac{1}{x}-\frac{1}{\sin (x)}=\frac{\sin (x)-x}{x\sin (x)}\sim \frac{-(1/6)x^3}{x^2}=-\frac{1}{6}x$ – Git Gud Mar 24 '13 at 13:04
  • @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. –  Mar 24 '13 at 13:25
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    @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.) – Did Mar 24 '13 at 13:47
  • @Did I guessed the combining part was left to me which I did and found it worked. –  Mar 24 '13 at 13:50
  • @Stephen Good for you. But do not say "the argument combines Taylor series in an appropriate way" when there is no such thing there. – Did Mar 24 '13 at 13:55
  • @Did Did you attempt that step? –  Mar 24 '13 at 13:57
  • @Stephen Sorry? Are you asking me if I can solve the exercise using the expansions of $\csc^2$ and $\cot$ in Sami's post? – Did Mar 24 '13 at 13:59
  • @Did I'm simply saying that the step is quite easy so I don't understand the problem –  Mar 24 '13 at 14:01
  • As I said, what you call "the step" is roughly as easy or as difficult as the exercise itself. But the ways of the mind are mysterious, aren't they? Hence, if indeed this indication led you to a full solution, then everything is fine. – Did Mar 24 '13 at 14:03
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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.

Mikasa
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    You don't have to apologize. I think your answer suffers from the same issues as Sami's. – Git Gud Mar 24 '13 at 20:56
  • @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.:) – Mikasa Mar 24 '13 at 20:59
  • @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. –  Mar 24 '13 at 23:43
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    @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. – Git Gud Mar 24 '13 at 23:47