Is substitution a valid approach?

The Maclaurin series for the two trig functions are

$\tan(x) = x + \dfrac{x^{3}}{3} + \dfrac{2x^{5}}{15} + O(x^{7})$ and

$\sin(x) = x - \dfrac{x^{3}}{6} + \dfrac{x^{5}}{120} - O(x^{7})$ .

So $\dfrac{\tan(x) - x}{x - \sin(x)} = \dfrac{\dfrac{x^{3}}{3} + \dfrac{2x^{5}}{15} + O(x^{7})}{\dfrac{x^{3}}{6} - \dfrac{x^{5}}{120} + O(x^{7})} = \dfrac{\dfrac{1}{3} + O(x^{2})}{\dfrac{1}{6} + O(x^{2})}$ ,

where in this last step I divided the numerator and denominator through by $x^{3}$ . Now in the limit as $x \to 0$ the $O(x^{2})$ terms go to $0$ , leaving us with

$\displaystyle\lim_{x \to 0} \dfrac{\tan(x) - x}{x - \sin(x)} = \dfrac{\dfrac{1}{3}}{\dfrac{1}{6}} = \boxed{2}$ .

You can apply L'Hopital's rule twice.

$=\lim_{x \rightarrow 0} \frac{\sec^{2}x - 1}{1-\cos x}$

This still is in the form $\frac{0}{0}$ , so we are allowed to reapply L'Hopital's.

$= \lim_{x \rightarrow 0} \frac{2\sec^{2} x \tan x }{\sin x}$

This can be rewritten as

$= \lim_{x \rightarrow 0} \frac{2}{\cos^{3} x}$

using the identities $\frac{\sin x}{\cos x} = \tan x$ and $\sec x = \frac{1}{\cos x}$ .

Finally, substitute in $x = 0$ . Since $\cos 0 = 1$ , the answer is $\boxed{2}$ .