I want to evaluate $\lim\limits_{x\to 0} \frac{x(1-\cos x)}{x - \sin(x)}$. We know that its plot is:
And also with attention to the Taylor series we know that its series expansion at $x=0$ is:
$$3 - x^2/10 - x^4/4200 + O(x^6)$$
So the limit is $3$. But I want to evaluate it without Taylor series or L'Hôpital's rule.
What I tried: $$\begin{aligned} \lim\limits_{x\to 0} \frac{x(1-\cos x)}{x - \sin(x)}& =\\ &\lim_{x\to 0}\frac{x\cdot2\sin^2\frac{x}{2}}{x - \sin x} = \lim_{x\to 0}\frac{2x\cdot\sin^2\frac{x}{2}}{x\cdot(1 - \frac{\sin x}{x})} = \\&\lim_{x\to 0}\frac{2\sin^2\frac{x}{2}}{1 - \frac{\sin x}{x}} \end{aligned}$$ But that did not help to evaluate the indeterminate form. Also I want to know is there a geometric representation for $x - \sin(x)$ which helps us to evaluate the limit?