TITLE: A limit problem $\lim\limits_{x \to 0}\frac{x\sin(\sin x) - \sin^{2}x}{x^{6}}$
QUESTION [32 upvotes]: 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 $x = 0.01$

REPLY [5 votes]: 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?