<s>[INST] Limit of $\frac{1}{x}\left( {1 - \frac{{\sin x}}{x}} \right)$ $\mathop {\lim }\limits_{x \to 0} \frac{1}{x}\left( {1 - \frac{{\sin x}}{x}} \right) = 0$
The result can be checked with Taylor series or with L'Hôpital's rule.
I wonder if it's possible to reach the same result with some standard algebraic manipulations (given that $\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x} = 1$)
 [/INST] Following @labbhattacharjee hint here's my elaboration on Are all limits solvable without L'Hôpital Rule or Series Expansion.
I've added this "self-reply" in the hope it might be useful for others in this specific case. I've also added some more detail to the linked post derivation.
$L_0=\mathop {\lim }\limits_{x \to 0} \frac{1}{x}\left( {1 - \frac{{\sin x}}{x}} \right) = \mathop {\lim }\limits_{x \to 0} \frac{{x - \sin x}}{{{x^3}}}x$
We calculate the limit $L$ of the first factor inside $L_0$
$L = \mathop {\lim }\limits_{x \to 0} \frac{{x - \sin x}}{{{x^3}}} = \mathop {\lim }\limits_{y \to 0} \frac{{3y - \sin 3y}}{{27{y^3}}}$ where we have set $y = 3x$
Now we must use the trigonometric identity
$\sin 3y = \sin \left( {2y + y} \right) = \sin 2y\cos y + \cos 2y\sin y = 2\sin y{\cos ^2}y + \left( {1 - 2{{\sin }^2}y} \right)\sin y = 2\sin y - 2{\sin ^3}y + \sin y - 2{\sin ^3}y$
that is 
$\sin 3y = 3\sin y - 4{\sin ^3}y$
Then we have
$L = \mathop {\lim }\limits_{y \to 0} \frac{{3y - \sin 3y}}{{27{y^3}}} = \mathop {\lim }\limits_{y \to 0} \frac{{3y - 3\sin y + 4{{\sin }^3}y}}{{27{y^3}}} = \mathop {\lim }\limits_{y \to 0} \frac{{3y - 3\sin y}}{{27{y^3}}} + \mathop {\lim }\limits_{y \to 0} \frac{{4{{\sin }^3}y}}{{27{y^3}}} = \frac{1}{9}\mathop {\lim }\limits_{y \to 0} \frac{{y - \sin y}}{{{y^3}}} + \frac{4}{{27}}$
$L = \mathop {\lim }\limits_{x \to 0} \frac{{x - \sin x}}{{{x^3}}} = \mathop {\lim }\limits_{y \to 0} \frac{{3y - \sin 3y}}{{27{y^3}}} = \mathop {\lim }\limits_{y \to 0} \frac{{y - \sin y}}{{{y^3}}} = \frac{1}{9}L + \frac{4}{{27}}$
So we have the equation
$L = \frac{1}{9}L + \frac{4}{{27}}$
whose solution is
$L = \frac{1}{6}$
The original limit is then $L_0=0$ as it is the product of a finite number and an infinitesimal quantity
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