<s>[INST] 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 $
 [/INST] 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 $$\lim_{x \rightarrow 0} \frac{-(\sin(x) -x)}{x^2} = \lim_{x \rightarrow 0} -(\frac{-x^3/3!}{x^2} )= \lim_{x \rightarrow 0} x/6 = 0$$
</s>