Explain how to solve this trigonometric limit without L'Hôpital's rule?

Question

In my previous class our professor let us the following limit:

$$ \lim_{x\to0} \frac{\tan(x)-\sin(x)}{x-\sin(x)} $$

He solved it by applying L'Hôpital's rule as follow:

$ \lim_{x\to0} \frac{\sec^2(x) - \cos(x)}{1-\cos(x)} = \lim_{x\to0} \frac{\cos^2(x) + \cos(x) + 1}{\cos^2(x)} = \frac{3}{1} = 3 $

He only wrote that in the blackboard and then told us to solve it but without using L'Hôpital's rule. So I proceeded in this way:

$ \lim_{x\to0} \frac{\frac{\sin(x)}{\cos(x)} - \sin(x)}{x - \sin(x)} = \lim_{x\to0} \frac{\frac{\sin(x)(1 - \cos(x))}{\cos(x)}}{x - \sin(x)} = \lim_{x\to0} \frac{\sin(x)(1 - \cos(x))}{(x - \sin(x))(\cos(x))} = \lim_{x\to0} \frac{\sin(x)(1 - \cos(x))}{(x - \sin(x))(\cos(x))} $

From there I multiply by its conjugate $(1 - \cos(x))$ but then I get more confused:

$ \lim_{x\to0} \frac{\sin(x)(1 - \cos(x))}{(x - \sin(x))(\cos(x))} \frac{(1 + \cos(x))}{(1 + \cos(x))} = \lim_{x\to0} \frac{\sin^3(x)}{(x - \sin(x))(\cos(x))(1 + \cos(x))} $ ...

Can someone give me a better advice on how to get the right result.

Answer 1

HINT

By Taylor's expansion

$$ \frac{\tan(x)-\sin(x)}{x-\sin(x)}= \frac{x+\frac{x^3}3-x+\frac{x^3}6+o(x^3)}{x-x+\frac{x^3}6+o(x^3)}=\frac{\frac{x^3}2+o(x^3)}{\frac{x^3}6+o(x^3)}=\frac{\frac{1}2+o(1)}{\frac{1}6+o(1)}\to 3$$

Answer 2

$$\dfrac{\tan x-\sin x}{x-\sin x}=\dfrac1{\cos x}\cdot\dfrac{\sin x}x\cdot\dfrac{1-\cos x}{x^2}\cdot\dfrac1{\dfrac{x-\sin x}{x^3}}$$

$\dfrac{1-\cos x}{x^2}=\left(\dfrac{\sin x}x\right)^2\cdot\dfrac1{1+\cos x}$

Finally for $$\dfrac{x-\sin x}{x^3}$$ use Are all limits solvable without L'Hôpital Rule or Series Expansion