A limit problem $\lim\limits_{x \to 0}\frac{x\sin(\sin x) - \sin^{2}x}{x^{6}}$ 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 $x = 0.01$
 A: $\eqalign{
L
\;&=\;
\lim\limits_{x \to 0}   \dfrac     {x-\sin x}     {x^3}
\\
\;&=\;
\lim\limits_{t \to 0}    \dfrac   {3t-\sin 3t}    {27t^3}\qquad x=3t
\\
\;&=\;
\lim\limits_{t \to 0}   \dfrac{3t-3\sin t+4\sin^3t}{27t^3}\qquad \sin3t=3\sin  t-4\sin^3t
\\
\;&=\;
\dfrac19\lim\limits_{t \to 0}   \dfrac{t-\sin t}{t^3}+\dfrac4{27}\lim\limits_{t\to0}\dfrac{\sin^3t}{t^3}
\\
\;&=\;
\dfrac L9+\dfrac4{27}
\\
\;&\Rightarrow\;\;\,\text{we obtain }\, L=\dfrac16
}$
A: 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}
A: 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 $\lim\limits_{x\to0}\dfrac{x \sin(s) - s^2}{x^6}=\dfrac1{12}-\dfrac1{36}=\dfrac1{18}$.
A: 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.
A: 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*}
