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
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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