A limit problem $\lim\limits_{x \to 0}\frac{x\sin(\sin x) - \sin^{2}x}{x^{6}}$

Question

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 $x = 0.01$

Answer 1

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, $$ \frac{\sin(A)-\sin(B)}{\sin(A-B)} =\frac{\cos\left(\frac{A+B}{2}\right)}{\cos\left(\frac{A-B}{2}\right)} =1-\frac{2\sin\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)}{\cos\left(\frac{A-B}{2}\right)}\tag{11} $$

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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 $$ \color{#C00000}{\lim_{x\to0}\frac{x\sin(\sin(x))-\sin^2(x)}{x^6}=\frac1{18}}\tag{15} $$

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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 $(6)$.

Answer 2

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}

Answer 3

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 $\lim\limits_{x\to0}\dfrac{x \sin(s) - s^2}{x^6}=\dfrac1{12}-\dfrac1{36}=\dfrac1{18}$.

Answer 4

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.

Answer 5

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