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This is a problem from "A Course of Pure Mathematics" by G H Hardy. Find the limit $$\lim_{x \to 0}\frac{x\sin(\sin x) - \sin^{2}x}{x^{6}}$$ I had solved it long back (solution presented in my blog here) but I had to use the L'Hospital's Rule (another alternative is Taylor's series). This problem is given in an introductory chapter on limits and the concept of Taylor series or L'Hospital's rule is provided in a later chapter in the same book. So I am damn sure that there is a mechanism to evaluate this limit by simpler methods involving basic algebraic and trigonometric manipulations and use of limit $$\lim_{x \to 0}\frac{\sin x}{x} = 1$$ but I have not been able to find such a solution till now. If someone has any ideas in this direction please help me out.

PS: The answer is $1/18$ and can be easily verified by a calculator by putting $x = 0.01$

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@ParamanandSingh: Did they do series expansions early on? $f(x) = 1/18-x^2/45+(569 x^4)/113400+O(x^6)$ – Amzoti Jul 7 '13 at 5:45
No. It's easy if we could use both $\sin(s) = s - \dfrac{s^3}6 + \dfrac{s^5}{120} + o(s^6)$ and $x = \arcsin(s) = s + \dfrac{s^3}6 + \dfrac{3s^5}{40} + o(s^6)$. Let $s=\sin(x)$. Then \begin{align*} \lim_{x\to 0}\frac{x\sin(s)-s^2}{x^6} &=\lim_{x\to 0}\frac{(x-s)(\sin(s)-s)+s\left(\sin(s)-s\right)+s(x-s)}{x^6}\\ &=\lim_{x\to 0}\frac{ \left(\dfrac{s^3}6\right)\left(-\dfrac{s^3}6\right) + s\left(-\dfrac{s^3}6+\dfrac{s^5}{120}\right) + s\left(\dfrac{s^3}6+\dfrac{3s^5}{40}\right)}{x^6}\\ &=\lim_{x\to 0}\frac{\dfrac{-s^6}{36}+\dfrac{s^6}{120}+\dfrac{3s^6}{40}}{x^6} =\frac1{18}. \end{align*} – user1551 Jul 7 '13 at 13:13
Out of curosity, where exactly is this problem found in the book? I've been trying to find it in my copy, but to no avail. – Andrew D Jul 7 '13 at 19:30
Thanks to all the people who put in effort for solving the problem. I particularly like the detailed solution by robjohn which establishes the fundamental limit of $(x - \sin x)/x^{3}$ without going through L'Hospital. Although it is bit detailed, it seems to be the best. As to the comment from Andrew D, this is from 10th edition of the book and it asks "Prove that $\phi(x) = x\sin(\sin x) - \sin^{2}x$ is of the sixth order of smallness when $x$ is small; and find the limit of $\phi(x)/x^{6}$ as $x \to 0$" (check page no 207) – Paramanand Singh Jul 8 '13 at 3:54
up vote 29 down vote accepted

Preliminary Results:

We will use $$ \begin{align} \frac{\color{#C00000}{\sin(2x)-2\sin(x)}}{\color{#00A000}{\tan(2x)-2\tan(x)}} &=\underbrace{\color{#C00000}{2\sin(x)(\cos(x)-1)}\vphantom{\frac{\tan^2(x)}{\tan^2(x)}}}\underbrace{\frac{\color{#00A000}{1-\tan^2(x)}}{\color{#00A000}{2\tan^3(x)}}}\\ &=\hphantom{\sin}\frac{-2\sin^3(x)}{\cos(x)+1}\hphantom{\sin}\frac{\cos(x)\cos(2x)}{2\sin^3(x)}\\ &=-\frac{\cos(x)\cos(2x)}{\cos(x)+1}\tag{1} \end{align} $$ Therefore, $$ \lim_{x\to0}\frac{\sin(x)-2\sin(x/2)}{\tan(x)-2\tan(x/2)}=-\frac12\tag{2} $$ Thus, given an $\epsilon\gt0$, we can find a $\delta\gt0$ so that if $|x|\le\delta$ $$ \left|\,\frac{\sin(x)-2\sin(x/2)}{\tan(x)-2\tan(x/2)}+\frac12\,\right|\le\epsilon\tag{3} $$ Because $\,\displaystyle\lim_{x\to0}\frac{\sin(x)}{x}=\lim_{x\to0}\frac{\tan(x)}{x}=1$, we have $$ \sin(x)-x=\sum_{k=0}^\infty2^k\sin(x/2^k)-2^{k+1}\sin(x/2^{k+1})\tag{4} $$ and $$ \tan(x)-x=\sum_{k=0}^\infty2^k\tan(x/2^k)-2^{k+1}\tan(x/2^{k+1})\tag{5} $$ By $(3)$ each term of $(4)$ is between $-\frac12-\epsilon$ and $-\frac12+\epsilon$ of the corresponding term of $(5)$. Therefore, $$ \left|\,\frac{\sin(x)-x}{\tan(x)-x}+\frac12\,\right|\le\epsilon\tag{6} $$ Thus, $$ \lim_{x\to0}\,\frac{\sin(x)-x}{\tan(x)-x}=-\frac12\tag{7} $$ Furthermore, $$ \begin{align} \frac{\tan(x)-\sin(x)}{x^3} &=\tan(x)(1-\cos(x))\frac1{x^3}\\ &=\frac{\sin(x)}{\cos(x)}\frac{\sin^2(x)}{1+\cos(x)}\frac1{x^3}\\ &=\frac1{\cos(x)(1+\cos(x))}\left(\frac{\sin(x)}{x}\right)^3\tag{8} \end{align} $$ Therefore, $$ \lim_{x\to0}\frac{\tan(x)-\sin(x)}{x^3}=\frac12\tag{9} $$ Combining $(7)$ and $(9)$ yield $$ \lim_{x\to0}\frac{x-\sin(x)}{x^3}=\frac16\tag{10} $$ Additionally, $$ \frac{\sin(A)-\sin(B)}{\sin(A-B)} =\frac{\cos\left(\frac{A+B}{2}\right)}{\cos\left(\frac{A-B}{2}\right)} =1-\frac{2\sin\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)}{\cos\left(\frac{A-B}{2}\right)}\tag{11} $$

Finishing Up: $$ \begin{align} &x\sin(\sin(x))-\sin^2(x)\\ &=[\color{#C00000}{(x-\sin(x))+\sin(x)}][\color{#00A000}{(\sin(\sin(x))-\sin(x))+\sin(x)}]-\sin^2(x)\\ &=\color{#C00000}{(x-\sin(x))}\color{#00A000}{(\sin(\sin(x))-\sin(x))}\\ &+\color{#C00000}{(x-\sin(x))}\color{#00A000}{\sin(x)}\\ &+\color{#C00000}{\sin(x)}\color{#00A000}{(\sin(\sin(x))-\sin(x))}\\ &=(x-\sin(x))(\sin(\sin(x))-\sin(x))+\sin(x)(x-2\sin(x)+\sin(\sin(x)))\tag{12} \end{align} $$ Using $(10)$, we get that $$ \begin{align} &\lim_{x\to0}\frac{(x-\sin(x))(\sin(\sin(x))-\sin(x))}{x^6}\\ &=\lim_{x\to0}\frac{x-\sin(x)}{x^3}\lim_{x\to0}\frac{\sin(\sin(x))-\sin(x)}{\sin^3(x)}\lim_{x\to0}\left(\frac{\sin(x)}{x}\right)^3\\ &=\frac16\cdot\frac{-1}6\cdot1\\ &=-\frac1{36}\tag{13} \end{align} $$ and with $(10)$ and $(11)$, we have $$ \begin{align} &\lim_{x\to0}\frac{\sin(x)(x-2\sin(x)+\sin(\sin(x)))}{x^6}\\ &=\lim_{x\to0}\frac{\sin(x)}{x}\lim_{x\to0}\frac{x-2\sin(x)+\sin(\sin(x))}{x^5}\\ &=\lim_{x\to0}\frac{(x-\sin(x))-(\sin(x)-\sin(\sin(x))}{x^5}\\ &=\lim_{x\to0}\frac{(x-\sin(x))-\sin(x-\sin(x))\left(1-\frac{2\sin\left(\frac{x}{2}\right)\sin\left(\frac{\sin(x)}{2}\right)}{\cos\left(\frac{x-\sin(x)}{2}\right)}\right)}{x^5}\\ &=\lim_{x\to0}\frac{(x-\sin(x))-\sin(x-\sin(x))+\sin(x-\sin(x))\frac{2\sin\left(\frac{x}{2}\right)\sin\left(\frac{\sin(x)}{2}\right)}{\cos\left(\frac{x-\sin(x)}{2}\right)}}{x^5}\\ &=\lim_{x\to0}\frac{\sin(x-\sin(x))}{x^3}\frac{2\sin\left(\frac{x}{2}\right)\sin\left(\frac{\sin(x)}{2}\right)}{x^2}\\[6pt] &=\frac16\cdot\frac12\\[6pt] &=\frac1{12}\tag{14} \end{align} $$ Adding $(13)$ and $(14)$ gives $$ \color{#C00000}{\lim_{x\to0}\frac{x\sin(\sin(x))-\sin^2(x)}{x^6}=\frac1{18}}\tag{15} $$

Added Explanation for the Derivation of $(6)$

The explanation below works for $x\gt0$ and $x\lt0$. Just reverse the red inequalities.

Assume that $x\color{#C00000}{\gt}0$ and $|x|\lt\pi/2$. Then $\tan(x)-2\tan(x/2)\color{#C00000}{\gt}0$.

$(3)$ is equivalent to $$ \begin{align} &(-1/2-\epsilon)(\tan(x)-2\tan(x/2))\\[4pt] \color{#C00000}{\le}&\sin(x)-2\sin(x/2)\\[4pt] \color{#C00000}{\le}&(-1/2+\epsilon)(\tan(x)-2\tan(x/2))\tag{16} \end{align} $$ for all $|x|\lt\delta$. Thus, for $k\ge0$, $$ \begin{align} &(-1/2-\epsilon)(2^k\tan(x/2^k)-2^{k+1}\tan(x/2^{k+1}))\\[4pt] \color{#C00000}{\le}&2^k\sin(x/2^k)-2^{k+1}\sin(x/2^{k+1})\\[4pt] \color{#C00000}{\le}&(-1/2+\epsilon)(2^k\tan(x/2^k)-2^{k+1}\tan(x/2^{k+1}))\tag{17} \end{align} $$ Summing $(17)$ from $k=0$ to $\infty$ yields $$ \begin{align} &(-1/2-\epsilon)\left(\tan(x)-\lim_{k\to\infty}2^k\tan(x/2^k)\right)\\[4pt] \color{#C00000}{\le}&\sin(x)-\lim_{k\to\infty}2^k\sin(x/2^k)\\[4pt] \color{#C00000}{\le}&(-1/2+\epsilon)\left(\tan(x)-\lim_{k\to\infty}2^k\tan(x/2^k)\right)\tag{18} \end{align} $$ Since $\lim\limits_{k\to\infty}2^k\tan(x/2^k)=\lim\limits_{k\to\infty}2^k\sin(x/2^k)=x$, $(18)$ says $$ \begin{align} &(-1/2-\epsilon)(\tan(x)-x)\\[4pt] \color{#C00000}{\le}&\sin(x)-x\\[4pt] \color{#C00000}{\le}&(-1/2+\epsilon)(\tan(x)-x))\tag{19} \end{align} $$ which, since $\epsilon$ is arbitrary is equivalent to $(6)$.

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(+1) Side note: it is easier to reinvent L'Hopitale's rule than come up with this solution. I bet that authors of the book didn't bother much about methods avaliable to students when they were suggesting this problem – Norbert Jul 7 '13 at 14:48
@Norbert It is rather naive to say that (essentially, sloppy or careless) about G. H. Hardy. – Andrés E. Caicedo Jul 7 '13 at 15:03
You like challenge, don't you? I think Hardy made a joke, here. – 1015 Jul 7 '13 at 15:43
@AndresCaicedo I will not change my mind even if now I realized that the author of the book is G. H. Hardy. I saw quite enough problems in books of well known mathematicians which far away from the level of other problem of a particular paragraph. – Norbert Jul 7 '13 at 16:03
The colors, the formatting, everything: it's just so beautiful. (+1000) if I could. – Omnomnomnom Jul 10 '13 at 15:33

Lemma. $\lim\limits_{x\to0}\dfrac{x-\sin(x)}{x^3}=\dfrac16$.

Proof of lemma. To prove this lemma, we will mainly follow robjohn's idea, but using a different proof. Since $\dfrac{x-\sin(x)}{x^3}$ is an even function, it suffices to prove that the right hand limit is equal to $\frac16$. For any fixed $0<x<\frac\pi2$, let $x_n = 2^{-n}x$ for $k=0,1,2,\ldots$. Then $$ \dfrac{\sin x_n}{x_n}=\frac{\sin(2x_{n+1})}{2x_{n+1}}=\frac{2\sin(x_{n+1})\cos(x_{n+1})}{2x_{n+1}}\le\frac{\sin(x_{n+1})}{x_{n+1}}. $$ So, $\color{red}{y_n} = \dfrac{\sin x_n}{x_n}$ is an increasing sequence. Now \begin{align*} \frac{\sin(x)-x}{x^3} &= \sum_{k=0}^n \frac{2^k\sin(x_k)-2^{k+1}\sin(x_{k+1})}{x^3} + \frac{2^{n+1}\sin(x_{n+1})-x}{x^3}\\ &= \sum_{k=0}^n \frac{2^{k+1}\sin(x_{k+1})\cos(x_{k+1})-2^{k+1}\sin(x_{k+1})}{x^3} + \frac{2^{n+1}\sin(x_{n+1})-x}{x^3}\\ &= -\sum_{k=0}^n \frac{2^{k+2}\sin(x_{k+1})\sin^2(x_{k+2})}{x^3} + \frac{\sin(x_{n+1})/x_{n+1} - 1}{x^2}\\ &= -\sum_{k=0}^n \frac{y_{k+1}y_{k+2}^2}{2^{2k+3}} + \frac{\sin(x_{n+1})/x_{n+1} - 1}{x^2}. \end{align*} Therefore \begin{align*} \frac{\sin(x)-x}{x^3} \begin{cases} \ge -\sum_{k=0}^n \frac{1}{2^{2k+3}} + \frac{\sin(x_{n+1})/x_{n+1} - 1}{x^2},\\ \le -\sum_{k=0}^n \frac{y_1^3}{2^{2k+3}} + \frac{\sin(x_{n+1})/x_{n+1} - 1}{x^2}. \end{cases}\tag{1} \end{align*} As $\sum_{k=0}^\infty \frac{1}{2^{2k+3}} = \frac16$, by taking $n$ to infinity in $(1)$, we get $$ -\frac16 \le \frac{\sin(x)-x}{x^3} \le -\frac{y_1^3}6. $$ Let $x\to0^+$, the result follows.

We are now ready to answer the OP's question.

Solution. Let $s=\sin(x)$. We have \begin{align*} &x \sin(s) - s^2\\ =&x \sin(s-x+x) - s^2\\ =&x \sin(s-x)\cos(x) + x\sin(x)\cos(s-x) - s^2\\ =&-x \sin(x-s)\cos(x) + xs \cos(x-s) - s^2\\ =&-x \sin(x-s)\cos(x) + xs - s^2 - xs(1 - \cos(x-s))\\ =&-x \sin(x-s)\cos(x) + x(x-s) - (x-s)^2 - xs(1 - \cos(x-s))\\ =&\underbrace{x((x-s)-\sin(x-s))\cos(x)}_A + \underbrace{x(x-s)(1-\cos(x))}_B - \underbrace{(x-s)^2}_C - \underbrace{xs(1 - \cos(x-s))}_D\\ =&A + B - C - D. \end{align*} Now \begin{align*} \lim_{x\to0}\frac{A}{x^6} &=\lim_{x\to0}\frac{(x-s)-\sin(x-s)}{x^5} =\lim_{x\to0}\frac{(x-s)-\sin(x-s)}{(x-s)^3}\left(\frac{x-s}{x^3}\right)^3 x^4=0,\\ \lim_{x\to0}\frac{B}{x^6} &=\lim_{x\to0}\frac{(x-s)(1-\cos(x))}{x^5} =\lim_{x\to0}\frac{x-s}{x^3}\frac{2\sin^2(x/2)}{x^2} =\frac16\times 2(1/2)^2 = \frac1{12},\\ \lim_{x\to0}\frac{C}{x^6} &=\lim_{x\to0}\frac{(x-s)^2}{x^6} =\left(\frac16\right)^2 = \frac1{36},\\ \lim_{x\to0}\frac{D}{x^6} &=\lim_{x\to0}\frac{1 - \cos(x-s)}{x^4} =\lim_{x\to0}\frac{2\sin^2(\frac{x-s}2)}{x^4} =\lim_{x\to0}\frac{2\sin^2(\frac{x-s}2)}{(x-s)^2}\left(\frac{x-s}{x^3}\right)^2x^2 =0. \end{align*} Therefore $\lim\limits_{x\to0}\dfrac{x \sin(s) - s^2}{x^6}=\dfrac1{12}-\dfrac1{36}=\dfrac1{18}$.

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Your solution is particularly appealing once we get the limit $(x - \sin x)/x^{3}$. Way better than what I presented in my blog post. – Paramanand Singh Jul 8 '13 at 3:57
@ParamanandSingh Thanks. Your question is very interesting. I really enjoy this fun exercise. – user1551 Jul 8 '13 at 12:15

I could prove it without using L'Hospital's rule, though I needed the following formula for $\sin x$ $$\sin{x}=x\prod_{k=1}^\infty\left(1-\frac{x^2}{k^2\pi^2}\right)=x\left(1-\frac{x^2}{6}+\frac{x^4}{120}+O(x^6)\right)$$ and the observation $$\sin ^2x=x^2\left(1-\frac{x^2}{3}+O(x^4)\right)$$ The constants $1/6$ and $1/120$ are due to $\zeta(2)/\pi^2$ and $\frac{1}{2}(\zeta^2(2)-\zeta(4))$ respectively. I also have used the simple formula $$\lim_{x\rightarrow 0}\frac{\sin x}{x}=1$$

Now I start the proof \begin{equation} \begin{split} \ &\lim_{x\rightarrow 0}\frac{x\sin(\sin x)-\sin^2x}{x^6}\\ \ = & \lim_{x\rightarrow 0}\frac{x\sin x \left(\prod_{k=1}^\infty\left(1-\frac{\sin^2x}{k^2\pi^2}\right)\right)-\sin^2x}{x^6}\\ \ =&\lim_{x\rightarrow 0}\frac{\sin x}{x} \lim_{x\rightarrow 0}\frac{x \left(\prod_{k=1}^\infty\left(1-\frac{\sin^2x}{k^2\pi^2}\right)\right)-x\prod_{k=1}^\infty\left(1-\frac{x^2}{k^2\pi^2}\right)}{x^5}\\ \ =&\lim_{x\rightarrow 0}\frac{(1-\frac{\sin^2x}{6}+\frac{\sin^4 x}{120}+O(\sin^6x))-(1-\frac{x^2}{6}+\frac{x^4}{120}+O(x^6))}{x^4}\\ \ =&\lim_{x\rightarrow 0}\frac{1-\frac{\sin^2x}{x^2}}{6x^2}+\lim_{x\rightarrow 0}\frac{\sin^4x-x^4}{120x^4}\\ \ =&\lim_{x\rightarrow 0}\frac{1-(1-\frac{x^2}{3}+O(x^4))}{6x^2}+\frac{1}{120}(1-1)\\ \ =&\frac{1}{18}\hspace{0.6cm} \Box \end{split} \end{equation}

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There is nothing wrong in your derivation Samrat, but somehow your solution seems to be too complicated for a question in an introductory chapter on limits. I am searching for a solution which can be presented to a person who is learning limits and has the will to try challenging problems. – Paramanand Singh Jul 7 '13 at 9:35
Yes, I admit that to follow my derivation the person has to know the Euler's product formula for sin(x). But otherwise, it is just manipulation. Also you don't need to place those big-Oh notations there, I just put them for brevity. – Samrat Mukhopadhyay Jul 7 '13 at 9:39
Just before this problem in the book, there is a simpler limit $$\lim_{x \to 0}\frac{1 - \cos(1 - \cos x)}{x^{4}}$$ which can be easily solved if write $$1 - \cos(1 - \cos x) = 1 - \cos\left(2\sin^{2}\frac{x}{2}\right) = 2\sin^{2}\left(\sin^{2}\frac{x}{2}\right)$$ and I am trying to find a solution similar to this. – Paramanand Singh Jul 7 '13 at 10:23
yes, i also saw that. I was also trying to find something similar, but for this problem it was not working. So I thought I should post this solution, because it does not use L'Hospital. Well maybe there is a simple one, we have to carry on the search for that. – Samrat Mukhopadhyay Jul 7 '13 at 10:29
To compute this limit I need only the following basic limit $$ \lim_{x\rightarrow 0}\frac{1-\cos x}{x^{2}}=\frac{1}{2}. $$ Indeed$,$ I will re-write the original expression as a function of the expression of this basic limit as follows $$ \lim_{x\rightarrow 0}\frac{1-\cos (1-\cos x)}{x^{4}}=\lim_{x\rightarrow 0} \frac{1-\cos (1-\cos x)}{(1-\cos x)^{2}}\left( \frac{1-\cos x}{x^{2}}\right) ^{2}=\frac{1}{2}\left( \frac{1}{2}\right) ^{2}=\frac{1}{8}. $$ – Idris May 20 '15 at 7:01

$\eqalign{ L \;&=\; \lim\limits_{x \to 0} \dfrac {x-\sin x} {x^3} \\ \;&=\; \lim\limits_{t \to 0} \dfrac {3t-\sin 3t} {27t^3}\qquad x=3t \\ \;&=\; \lim\limits_{t \to 0} \dfrac{3t-3\sin t+4\sin^3t}{27t^3}\qquad \sin3t=3\sin t-\sin^3t \\ \;&=\; \dfrac19\lim\limits_{t \to 0} \dfrac{t-\sin t}{t^3}+\dfrac4{27}\lim\limits_{t\to0}\dfrac{\sin^3t}{t^3} \\ \;&=\; \dfrac L9+\dfrac4{27} \\ \;&\Rightarrow\;\;\,\text{we obtain }\, L=\dfrac16 }$

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I have been provoked and so must prove that $$\lim_{x\rightarrow0}\frac{x-\sin x}{x^3}=\frac16$$ With only my trusty Uru hammer and not with Taylor series, L'Hopital's rule, or infinite series. We have proved that $$\lim_{x\rightarrow0}\frac{\sin x}x=1$$ And we will also make extensive use of the double angle identity $$\tan \theta=\frac{2\tan\frac{\theta}2}{1-\tan^2\frac{\theta}2}$$ We have to push $2\tan\frac{\theta}2$ all the way out to $O(\tan^5\theta)$. $$\begin{align}2\tan\frac{\theta}2&=\tan\theta\left(1-\tan^2\frac{\theta}2\right)=\tan\theta-\frac14\tan^3\theta\left(1-\tan^2\frac{\theta}2\right)^2\\ &=\tan\theta-\frac14\tan^4\theta+\frac12\tan^3\theta\tan^2\frac{\theta}2-\frac14\tan^3\theta\tan^4\frac{\theta}2\\ &=\tan\theta-\frac14\tan^3\theta+\frac18\tan^5\theta\left(1-\tan^2\frac{\theta}2\right)^2-\frac14\tan^3\theta\tan^4\frac{\theta}2\\ &=\tan\theta-\frac14\tan^3\theta+\frac18\tan^5\theta-\frac14\tan^5\theta\tan^2\frac{\theta}2+\frac18\tan^5\theta\tan^4\frac{\theta}2-\frac14\tan^3\theta\tan^4\frac{\theta}2\\ &=\tan\theta-\frac14\tan^3\theta+\frac18\tan^5\theta-\frac{8\tan^7\frac{\theta}2}{\left(1-\tan^2\frac{\theta}2\right)^5}+\frac{4\tan^9\frac{\theta}2}{\left(1-\tan^2\frac{\theta}2\right)^5}-\frac{2\tan^7\frac{\theta}2}{\left(1-\tan^2\frac{\theta}2\right)^3}\\ &=\tan\theta-\frac14\tan^3\theta+\frac18\tan^5\theta+\frac{\tan^7\frac{\theta}2\left(-10+8\tan^2\frac{\theta}2-2\tan^4\frac{\theta}2\right)}{\left(1-\tan^2\frac{\theta}2\right)^5}\end{align}$$ We only offer summary results for $-\frac23\tan^3\frac{\theta}2$ and $\frac25\tan^5\frac{\theta}2$: $$-\frac23\tan^3\frac{\theta}2=-\frac1{12}\tan^3\theta+\frac1{16}\tan^5\theta+\frac{\tan^7\frac{\theta}2\left(-18+20\tan^2\frac{\theta}2-10\tan^4\frac{\theta}2+2\tan^6\frac{\theta}2\right)}{3\left(1-\tan^2\frac{\theta}2\right)^5}$$ $$\frac25\tan^5\frac{\theta}2=\frac1{80}\tan^5\theta+\frac{2\tan^7\frac{\theta}2\left(-5+10\tan^2\frac{\theta}2-10\tan^4\frac{\theta}2+5\tan^6\frac{\theta}2-\tan^8\frac{\theta}2\right)}{5\left(1-\tan^2\frac{\theta}2\right)^5}$$ The object of these expansions was that we have an expression which relates adjacent members of a sequence $$2\tan\frac{\theta}2-\frac23\tan^3\frac{\theta}2+\frac25\tan^5\frac{\theta}2=\tan\theta-\frac13\tan^3\theta+\frac15\tan^5\theta+\frac{\tan^7\frac{\theta}2\left(-270+280\tan^2\frac{\theta}2-140\tan^4\frac{\theta}2+40\tan^6\frac{\theta}2-3\tan^8\frac{\theta}2\right)}{15\left(1-\tan^2\frac{\theta}2\right)^5} =\tan\theta-\frac13\tan^3\theta+\frac15\tan^5\theta+\frac{2\tan^7\frac{\theta}2}{15}\left(\frac{-48}{\left(1-\tan^2\frac{\theta}2\right)^5}+\frac{-48}{\left(1-\tan^2\frac{\theta}2\right)^4}+\frac{-28}{\left(1-\tan^2\frac{\theta}2\right)^3}+\frac{-8}{\left(1-\tan^2\frac{\theta}2\right)^2}+\frac{-3}{\left(1-\tan^2\frac{\theta}2\right)}\right)$$ Since $0<1-\tan^2\frac{\theta}2<1$ it follows that $$0>2\tan\frac{\theta}2-\frac23\tan^3\frac{\theta}2+\frac25\tan^5\frac{\theta}2-\tan\theta+\frac13\tan^3\theta-\frac15\tan^5\theta>-18\tan^7\frac{\theta}2$$ For any $0<\theta<\frac{\pi}4$. Now we can start bisecting $$0>2^n\left(\tan\frac{\theta}{2^n}-\frac13\tan^3\frac{\theta}{2^n}+\frac15\tan^5\frac{\theta}{2^n}-\frac{\theta}{2^n}\right)-2^{n-1}\left(\tan\frac{\theta}{2^{n-1}}+\frac13\tan^3\frac{\theta}{2^{n-1}}-\frac15\tan^5\frac{\theta}{2^{n-1}}-\frac{\theta}{2^{n-1}}\right)>-9\cdot2^n\tan^7\frac{\theta}{2^n}>-\frac9{2^{6n}}\tan^7\theta=-\frac{9\tan^7\theta}{2^{6n}\left(1-\frac1{2^6}\right)}+\frac{9\tan^7\theta}{2^{6n+6}\left(1-\frac1{2^6}\right)}$$ Since all series are telescoping, we may sum them fron $n=1$ to $n=N$ and then $$0>2^N\left(\tan\frac{\theta}{2^N}-\frac13\tan^3\frac{\theta}{2^N}+\frac15\tan^5\frac{\theta}{2^N}-\frac{\theta}{2^N}\right)-\left(\tan\theta-\frac13\tan^3\theta+\frac15\tan^5\theta-\theta\right)>-\frac{\tan^7\theta}{7}+\frac{\tan^7\theta}{7\cdot2^{6N}}>-\frac17\tan^7\theta$$ Now we will take our first limit! $$\lim_{N\rightarrow\infty}2^N\left(\tan\frac{\theta}{2^N}-\frac13\tan^3\frac{\theta}{2^N}+\frac15\tan^5\frac{\theta}{2^N}-\frac{\theta}{2^N}\right)=\lim_{N\rightarrow\infty}\theta\frac{\tan\frac{\theta}{2^N}-\frac13\tan^3\frac{\theta}{2^N}+\frac15\tan^5\frac{\theta}{2^N}-\frac{\theta}{2^N}}{\frac{\theta}{2^N}}=\theta(1-0+0-1)=0$$ That in turn implies $$0\ge\theta-\tan\theta+\frac13\tan^3\theta-\frac15\tan^5\theta\ge-\frac17\tan^7\theta$$ Again for $0<\theta<\frac{\pi}4$, $$0\ge\frac{\theta-\tan\theta+\frac13\tan^3\theta-\frac15\tan^5\theta}{\theta^3}\ge-\frac{\tan^7\theta}{7\theta^3}$$ It follows by the squeeze theorem that $$\lim_{\theta\rightarrow0^+}\frac{\theta-\tan\theta+\frac13\tan^3\theta-\frac15\tan^5\theta}{\theta^3}=0$$ And since easily $$\lim_{\theta\rightarrow0^+}\frac{\tan^3\theta}{3\theta^3}=\frac13$$ And $$\lim_{\theta\rightarrow0^+}\frac{\tan^5\theta}{5\theta^3}=0$$ We must conclude that $$\lim_{\theta\rightarrow0^+}\frac{\theta-\tan\theta}{\theta^3}=-\frac13$$ Finally, $$\tan\theta=\frac{\sin\theta}{\cos\theta}=\sin\theta\left(\frac{1+\tan^2\frac{\theta}2}{1-\tan^2\frac{\theta}2}\right)=\sin\theta+\sin\theta\left(\frac{2\tan^2\frac{\theta}2}{1-\tan^2\frac{\theta}2}\right)$$ $$\lim_{\theta\rightarrow0^+}\frac{\theta-\sin\theta}{\theta^3}=\lim_{\theta\rightarrow0^+}\frac{\theta-\tan\theta+\frac{2\sin\theta\tan^2\frac{\theta}2}{1-\tan^2\frac{\theta}2}}{\theta^3}=\lim_{\theta\rightarrow0^+}\frac{\theta-\tan\theta}{\theta^3}+\lim_{\theta\rightarrow0^+}\frac{\frac{\sin\theta}{\theta}\left(\frac{\tan\frac{\theta}2}{\frac{\theta}2}\right)^2}{2\left(1-\tan^2\frac{\theta}2\right)}=-\frac13+\frac{(1)(1)^2}{2(1-0)}=\frac16$$ OK, maybe not the prettiest proof, but what can you do when you are provoked like that?

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Although it is not prettiest (as you mention) it is a very good solution to a difficult problem. +1. I wish someone should have provoked you much earlier. Thanks a lot for this wonderful answer!!! – Paramanand Singh May 17 at 9:25

For an elementary proof, I’m sure those that have been given are pretty much what Hardy had in mind. But if you want to use the Taylor (Maclaurin) expansion of the sine, then it’s really easy. The function $\sin\circ\sin$ has the expansion $x-x^3/3+x^5/10$, ignoring terms of degree $7$ and higher; this is perfectly easy to do by hand. And the expansion of $\sin^2$ is $x^2-x^4/3+2x^6/45$, even easier. The first term in the desired difference is $x^6/18$, and there you are.

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Lubin, I agree fully with what you say. It would have been better if we were allowed to use series expansions. But as I said the book expects this question to be solved using preliminary techniques. – Paramanand Singh Jul 8 '13 at 4:10

Here, I will provide an answer which do not involve any use of Taylor series. Also it uses L'Hospital's Rule a minimum of times, and in order to prove some basic limits only which are the following \begin{eqnarray*} \lim_{x\rightarrow 0}\frac{\sin x-x}{x^{3}} &=&-\frac{1}{6},\ \ \ \ \ \ \ \ \ and\ \ \ \ \ \ \ \ \ \ \lim_{x\rightarrow 0}\frac{\sin ^{-1}x-x}{% x^{3}}=\frac{1}{6} \\ \lim_{x\rightarrow 0}\frac{\sin x-x+\frac{x^{3}}{6}}{x^{5}} &=&\frac{1}{120},% \ \ \ \ \ \ \ \ \ and\ \ \ \ \ \ \ \ \ \ \lim_{x\rightarrow 0}\frac{% \sin ^{-1}x-x-\frac{x^{3}}{6}}{x^{5}}=\frac{3}{40}. \end{eqnarray*} First, note that the simple change of variable $\sin x=t$ shows that \begin{eqnarray*} \lim_{x\rightarrow 0}\frac{x\sin (\sin x)-\sin ^{2}(x)}{x^{6}} &=&\lim_{x\rightarrow 0}\frac{x\sin (\sin x)-\sin ^{2}(x)}{\sin ^{6}x}\left( \frac{\sin x}{x}\right) ^{6} \\ &=&\lim_{t\rightarrow 0}\frac{\sin t\sin ^{-1}t-t^{2}}{t^{6}}. \end{eqnarray*} Thus, actually, it is equivalent to answer the following equivalent question: \begin{equation*} \lim_{t\rightarrow 0}\frac{\sin t\sin ^{-1}t-t^{2}}{t^{6}}. \end{equation*}

At present, some algebraic manipulations have to be used. But before, I will explain how they are performed. If you want to write the number $56$ as a function of the number $15$ you can write \begin{equation*} 56=15\times 3+rest \end{equation*} and you find the $rest$ as \begin{equation*} rest=56-15\times 3=56-45=11. \end{equation*} The number $3$ was chosen (in $15\times 3$) so that the rest $11$ would be smaller that $15.$ Let's go back to our expression \begin{equation*} \left( \frac{\sin t\sin ^{-1}t-t^{2}}{t^{6}}\right) \end{equation*} we want to write it as a product of some expressions of the basic limits cited above. Since the expression contains the product $\sin t\sin ^{-1}t$ and the denominator is $t^{6}$ so the natural choice would be this product $% \left( \frac{\sin t-t}{t^{3}}\right) \left( \frac{\sin ^{-1}t-t}{t^{3}}% \right) ,$ then we try to find the 'rest' as follows \begin{equation*} rest=\left( \frac{\sin t\sin ^{-1}t-t^{2}}{t^{6}}\right) -\left( \frac{\sin t-t}{t^{3}}\right) \left( \frac{\sin ^{-1}t-t}{t^{3}}\right) \end{equation*} which after easy computations one gets \begin{eqnarray*} rest &=&\left( \frac{\sin t\sin ^{-1}t-t^{2}}{t^{6}}\right) -\left( \frac{% \sin t-t}{t^{3}}\right) \left( \frac{\sin ^{-1}t-t}{t^{3}}\right) \\ &=&\frac{t\sin t+t\sin ^{-1}t-2t^{2}}{t^{6}}. \end{eqnarray*} Now we simplify by $t$ and try to write the result as a new function (maybe product) of the expressions of basic limits. Since $\sin t$ and $\sin ^{-1}t$ are linear in the last expression, I mean there is no exponent on either of them, so no product is permitted now, but we should add expressions containing linear terms of $\sin t$ and $\sin ^{-1}t.$ The natural choice is the following \begin{equation*} \frac{\sin t+\sin ^{-1}t-2t}{t^{5}}=\left( \frac{\sin t-t+\frac{t^{3}}{6}}{% t^{5}}\right) +\left( \frac{\sin ^{-1}t-t-\frac{t^{3}}{6}}{t^{5}}\right) . \end{equation*} Now we resume the resulting computations as follows: \begin{eqnarray*} \left( \frac{\sin t\sin ^{-1}t-t^{2}}{t^{6}}\right) &=&\left( \frac{\sin t-t% }{t^{3}}\right) \left( \frac{\sin ^{-1}t-t}{t^{3}}\right) +\frac{t\sin t+t\sin ^{-1}t-2t^{2}}{t^{6}} \\ &=&\left( \frac{\sin t-t}{t^{3}}\right) \left( \frac{\sin ^{-1}t-t}{t^{3}}% \right) +\frac{\sin t+\sin ^{-1}t-2t}{t^{5}} \\ &=&\left( \frac{\sin t-t}{t^{3}}\right) \left( \frac{\sin ^{-1}t-t}{t^{3}}% \right) +\left( \frac{\sin t-t+\frac{t^{3}}{6}}{t^{5}}\right) +\left( \frac{% \sin ^{-1}t-t-\frac{t^{3}}{6}}{t^{5}}\right) \end{eqnarray*} Without the explanations given above these decompositions look tricky or very genus, as they fall down from the sky, but I hope I have provided sufficient details to make them natural.

At the end, passing to the limit as $t$ tends to $0$ and using the basic limits gives immediately \begin{equation*} \lim_{t\rightarrow 0}\left( \frac{\sin t\sin ^{-1}t-t^{2}}{t^{6}}\right) =\left( -\frac{1}{6}\right) \left( \frac{1}{6}\right) +\left( \frac{1}{120}% \right) +\left( \frac{3}{40}\right) =\frac{1}{18}.\ \blacksquare \end{equation*}

Now, permit me to bring to reader's attention that we have ONLY used BASIC limits, and not the functions involved themselves. Also, if we remove the extra explanations the answer is short.

Let $f$ and $g$ any two functions defined around $t=0$ such that \begin{eqnarray*} \lim_{t\rightarrow 0}\frac{f(t)-t}{t^{3}} &=&a,\ \ \ \ \ \ \ \ \ and\ \ \ \ \ \ \ \ \ \ \lim_{t\rightarrow 0}\frac{g(t)-t}{t^{3}}=b \\ \lim_{t\rightarrow 0}\frac{f(t)-t+\frac{t^{3}}{6}}{t^{5}} &=&c,\ \ \ \ \ \ \ \ \ and\ \ \ \ \ \ \ \ \ \ \lim_{t\rightarrow 0}\frac{g(t)-t-\frac{% t^{3}}{6}}{t^{5}}=d. \end{eqnarray*}

Then one can show that \begin{equation*} \lim_{t\rightarrow 0}\left( \frac{f(t)g(t)-t^{2}}{t^{6}}\right) =ab+c+d. \end{equation*}

PROOF (which is short). \begin{eqnarray*} \left( \frac{\mathsf{f(t)g(t)}-t^{2}}{t^{6}}\right) &=&\left( \frac{\mathsf{% f(t)}-t}{t^{3}}\right) \left( \frac{\mathsf{g(t)}-t}{t^{3}}\right) +\frac{t% \mathsf{f(t)}+t\mathsf{g(t)}-2t^{2}}{t^{6}} \\ &=&\left( \frac{\mathsf{f(t)}-t}{t^{3}}\right) \left( \frac{\mathsf{g(t)}-t}{% t^{3}}\right) +\frac{\mathsf{f(t)}+\mathsf{g(t)}-2t}{t^{5}} \\ &=&\left( \frac{\mathsf{f(t)}-t}{t^{3}}\right) \left( \frac{\mathsf{g(t)}-t}{% t^{3}}\right) +\left( \frac{\mathsf{f(t)}-t+\frac{t^{3}}{6}}{t^{5}}\right) +\left( \frac{\mathsf{g(t)}-t-\frac{t^{3}}{6}}{t^{5}}\right) \end{eqnarray*}

Passing to the limit when $t$ tends to $0$ and using the BASIC LIMITS given in the hypothesis one gets \begin{equation*} \lim_{t\rightarrow 0}\left( \frac{\mathsf{f(t)g(t)}-t^{2}}{t^{6}}\right) =ab+c+d.\ \blacksquare \end{equation*}

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Good explanation of the technique. +1 – Paramanand Singh May 20 '15 at 7:41
All the results one would use from Taylor series are in the four limits claimed at the beginning. – robjohn May 20 '15 at 8:14

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