Proof Check: $ \lim_{x \rightarrow 0} \frac{x-\sin x}{x^2} $ The limit can be rewritten as
$$ \lim_{x \rightarrow 0} \frac{x-\sin x}{x^2} = \lim_{x \rightarrow 0} \frac{1}{x} \left[ 1- \frac{\sin x}{x} \right] $$
Recall the inequality,
$$ \cos x < \frac{\sin x}{x} < 1$$
holds for $ x\in (-\pi/2, \pi/2) $.
This provides the new inequalities
$$ 0 < \frac{1}{x} \left[ 1- \frac{\sin x}{x} \right] < \frac{1-\cos x}{x},  \text{ with } x>0 $$
$$ \frac{1-\cos x}{x} < \frac{1}{x} \left[1- \frac{\sin x}{x} \right]< 0,  \text{ with } x<0 $$
Applying the Squeeze Theorem, we get
$$ \lim_{x \rightarrow 0^{\pm}}  \frac{1}{x} \left[ 1- \frac{\sin x}{x} \right] = 0$$
Note:
$ \lim_{ x \rightarrow 0} \frac{1-\cos x}{x} = 0 $
 A: Your proof is right. But I would have used Taylor developments :
$$\frac{x-\sin x}{x^2} \underset{x\rightarrow 0}{=} \frac{x-(x-x^3/6+o(x^3))}{x^2}\underset{x\rightarrow 0}{=}\frac{x}{6}+o(x)\underset{x\rightarrow 0}{\longrightarrow} 0$$
A: It is a fine and nice argument with trig functions.
Another way to proceed is to use asymptitics (see them as Taylor expansions at $x=0$.
Then you can write $$\lim_{x \rightarrow 0} \frac{-(\sin(x) -x)}{x^2} = \lim_{x \rightarrow 0} -(\frac{-x^3/3!}{x^2} )= \lim_{x \rightarrow 0} x/6 = 0$$
A: Here is another way to proceed it you don't wanna use L'Hopital, or series expansions.
Let $ x\in\mathbb{R}^{*} $, observe that : $$ \fbox{$\begin{array}{rcl}\displaystyle\frac{x-\sin{x}}{x^{2}}=\frac{x}{2}\int_{0}^{1}{\left(1-y\right)^{2}\cos{\left(xy\right)}\,\mathrm{d}y}\end{array}$} $$
Using the fact that $ \left(\forall t\in\mathbb{R}\right),\ \left|\cos{t}\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-y\right)^{2}\cos{\left(xy\right)}\,\mathrm{d}y}\right|&\leq\frac{\left|x\right|}{2}\int_{0}^{1}{\left(1-y\right)^{2}\left|\cos{\left(xy\right)}\right|\mathrm{d}y}\\&\leq\frac{\left|x\right|}{2}\int_{0}^{1}{\left(1-y\right)^{2}\,\mathrm{d}y}=\frac{\left|x\right|}{6}\underset{x\to 0}{\longrightarrow}0 \end{aligned}
Thus : $$ \lim\limits_{x\to 0}{\frac{x-\sin{x}}{x^{2}}}=0 $$
