Limit of $x^2(1 - x \sin \frac{1}{x})$ as $x \to \infty$ Just like in title, I want to find such limit:
$$
\lim_{x \to \infty}
\left\{x^{2}\left[1 - x\sin\left(1 \over x\right)\right]\right\}
$$
I just want to avoid using Taylor and stronger tools. It should be done propably using l'Hospital and maybe some neat identities or tricks. I've tried l'Hospital, but it just produces mess. Any hints would be appreciated.
 A: It's $$\lim_{x\rightarrow0}\frac{1-\frac{\sin{x}}{x}}{x^2}=\lim_{x\rightarrow0}\frac{x-\sin{x}}{x^3}=\lim_{x\rightarrow0}\frac{1-\cos{x}}{3x^2}=\lim_{x\rightarrow0}\frac{2\sin^2\frac{x}{2}}{3\cdot4\cdot\frac{x^2}{4}}=\frac{1}{6}.$$
Also, you can use $$\sin{x}=x-\frac{x^3}{6}+...$$ and we obtain $\frac{1}{6}$ immediately. 
A: with the help of the series Expansion for $x=\infty$ we get
$$\frac{1}{6}-\frac{1}{120x^2}+\frac{1}{5040x^4}+O\left(\left(\frac{1}{x}\right)^5\right)$$
A: We are dealing with limit $\dfrac{u-\sin u}{u^{3}}$ as $u\rightarrow 0$. But
\begin{align*}
\dfrac{u-\sin u}{u^{3}}&=\dfrac{1}{u^{3}}\int_{0}^{u}(1-\cos t)dt\\
&=\dfrac{1}{u^{3}}\int_{0}^{u}2\sin^{2}\left(\dfrac{t}{2}\right)dt.
\end{align*}
Given $\epsilon\in(0,1)$, for small $t>0$, then 
\begin{align*}
(1-\epsilon)\left(\dfrac{t}{2}\right)^{2}\leq\sin^{2}\left(\dfrac{t}{2}\right)\leq(1+\epsilon)\left(\dfrac{t}{2}\right)^{2},
\end{align*}
but
\begin{align*}
\dfrac{1}{u^{3}}\int_{0}^{u}2\left(\dfrac{t}{2}\right)^{2}dt=\dfrac{1}{6},
\end{align*}
so
\begin{align*}
\dfrac{1}{6}(1-\epsilon)\leq\dfrac{1}{u^{3}}\int_{0}^{u}2\sin^{2}\left(\dfrac{t}{2}\right)dt\leq\dfrac{1}{6}(1+\epsilon),
\end{align*}
this shows 
\begin{align*}
\dfrac{1}{u^{3}}\int_{0}^{u}2\sin^{2}\left(\dfrac{t}{2}\right)dt\rightarrow\dfrac{1}{6},~~~~u\rightarrow 0.
\end{align*}
