Computing limit of $frac{x-sin x}{x-tan x}$ without L'Hôpital

I want to compute following:

$$ \lim_{x \rightarrow 0} \frac{x-\sin(x)}{x-\tan(x)}$$

I have tried to calculate this with l’Hôpital's rule. l’Hôpital's rule states that:
$$ \lim_{x\rightarrow a} \frac{f(x)}{g(x)}=\lim_{x \rightarrow a } \frac{f'(x)}{g'(x)} $$

Now i can get to the right result with l’Hôpital's rule but it took a little over two pages on paper. Had to use l’Hôpital's rule 4 times. How do you solve this without l’Hôpital's rule ?

The result appears to be(with l’Hôpital's rule):

$$ \lim_{x \rightarrow 0} \frac{x-\sin(x)}{x-\tan(x)}=-\frac{1}{2} $$

If someone can provide alternative solution to this problem that would be highly appreciated.

Answer

Utilizing the Taylor expansions

$\sin(x)=x-\frac{1}{6}x^3+\mathcal{O}(x^5)$

$\tan(x)=x+\frac{1}{3}x^3+\mathcal{O}(x^5)$, we get

$\frac{x-\sin(x)}{x-\tan(x)}=\frac{x-(x-\frac{1}{6}x^3+\mathcal{O}(x^5))}{x-(x+\frac{1}{3}x^3+\mathcal{O}(x^5))}=\frac{\frac{1}{6}x^3+\mathcal{O}(x^5)}{-\frac{1}{3}x^3+\mathcal{O}(x^5)}\to-\frac{1}{2}$ as $x\to0$.