$\lim\limits_{x \to 0} \frac{\sin x-x}{x^2}$ without L'Hospital's rule $$\lim_{x \to 0} \frac{\sin x-x}{x^2}$$
I know it's an easy limit since inside the limit approaches $0/0$ as $x$ approaches $0$, we can use L'Hospital's twice to get 
$$\lim_{x \to 0} -\frac{\sin x}{2} = 0 $$
So my question is that: Is there any way to calculate the limit other than L'Hospital's rule ?
 A: Let $ x\in\mathbb{R}^{*} $, observe that : $$ \fbox{$\begin{array}{rcl}\displaystyle\frac{x-\sin{x}}{x^{2}}=\frac{x}{2}\int_{0}^{1}{\left(1-t\right)^{2}\cos{\left(tx\right)}\,\mathrm{d}t}\end{array}$} $$
Using the fact that $ \left(\forall t\in\left[0,1\right]\right),\ \left|\cos{\left(tx\right)}\right|\leq 1 $, we have : \begin{aligned} \left|\frac{x-\sin{x}}{x^{2}}\right|=\frac{\left|x\right|}{2}\left|\int_{0}^{1}{\left(1-t\right)^{2}\cos{\left(tx\right)}\,\mathrm{d}t}\right|&\leq\frac{\left|x\right|}{2}\int_{0}^{1}{\left(1-t\right)^{2}\left|\cos{\left(tx\right)}\right|\mathrm{d}t}\\&\leq\frac{\left|x\right|}{2}\int_{0}^{1}{\left(1-t\right)^{2}\,\mathrm{d}t} \end{aligned}
Which means $ \left(\forall x\in\mathbb{R}^{*}\right),\ \left|\frac{x-\sin{x}}{x^{2}}\right|\leq\frac{\left|x\right|}{6} $, and thus $ \lim\limits_{x\to 0}{\frac{x-\sin{x}}{x^{2}}}=0 \cdot $
A: $$
\lim_{x\to 0} \left|\frac{\sin(x)-x}{x^2} \right|= \lim_{x\to 0} \left|\frac{\frac{\sin(x)}{x}-1}{x} \right|
$$By some elementary geometric inequalities, for $x$ near $0$ we have
$$
\cos(x)\leq \frac{\sin(x)}{x}\leq 1
$$Thus
$$
  \lim_{x\to 0} \frac{1-1}{x}\leq \lim_{x\to 0}\left| \frac{\frac{\sin(x)}{x}-1}{x}\right| \leq  \lim_{x\to 0} \left|\frac{{1-\cos(x)}}{x}\right|
$$
$$
0\leq \lim_{x\to 0}\left| \frac{\frac{\sin(x)}{x}-1}{x}\right| \leq  \lim_{x\to 0} \left|\frac{\sin^2(x)}{x(1+\cos(x))}\right|
$$This last limit goes to $0$, since by the elementary inequality above, $\sin(x)/x\to 1$.
A: 
I thought it might be of interest to present a way forward that avoids using calculus and relies instead on pre-calculus tools only.  To that end we proceed.



We start with the trigonometric identity
$$\sin(3x)=3\sin(x)-4\sin^3(x) \tag 1$$
Next, we enforce the substitution $x\to x/3$ in $(1)$ yields
$$\sin(x)=3\sin(x/3)-4\sin^3(x/3)$$
Upon the subsequent iteration we obtain
$$\sin(x)=3^2\sin(x/3^2)-4\times 3^1\sin^3(x/3^2)-4\sin^3(x/3^1)$$ 
We have then after $n$ iterations
$$\sin(x)=3^n\sin(x/3^n)-4\sum_{k=1}^n3^{k-1}\sin^3(x/3^k) \tag 2$$
Using $\sin (x)\le x$ for $x>0$ in $(2)$ reveals
$$\sin(x)\ge 3^n\sin(x/3^n)-4x^3\sum_{k=1}^n 3^{k-1}/3^{3k}$$
Letting $n\to \infty$ yields
$$\sin(x)\ge x-\frac16x^3$$
for $x>0$.  Hence, for $x>0$ we see that 
$$-\frac16 x\le \frac{\sin(x)-x}{x^2}\le 0\tag3$$

Application of the squeeze theorem to $(3)$ reveals 
$$\lim_{x\to 0^+}\frac{\sin(x)-x}{x^2}=0$$
Now, use the analogous development to show that the limit from the left is $0$ to yield the coveted result
$$\lim_{x\to0}\frac{\sin(x)-x}{x^2}=0$$
And we are done!


Tools Used:  Trigonometric Series, Summation of a Geometric Series, The Squeeze Theorem

A: Yes: use the fact that$$(\forall x\in\Bbb R):\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots$$to deduce that$$(\forall x\in\Bbb R\setminus\{0\}):\frac{\sin(x)-x}{x^2}=-\frac x{3!}+\frac{x^3}{5!}-\cdots$$So, your limit is $0$.
