Calculate $\lim_{x\to 0} \frac{x-\sin x} {1-\cos x}$ Calculate the limit without using de l'Hopital:
$$\lim_{x\to 0} \frac{x-\sin x} {1-\cos x}$$ 
I want to use the limit:$$\lim_{x\to 0} \frac{\sin x}{x}=1$$ but I don't know how to do it.
I manipulated the expression to get
$$\lim_{x\to 0} \frac{x}{x-\sin x}-\lim_{x\to 0} \frac{\sin x} {1-\cos x}$$ 
but I don't know where to go from here.
 A: $f(x)=\frac{x-\sin(x)}{1-\cos(x)}$ is an odd function, hence as soon as we prove that $f(x)$ is continuous at the origin we have $\lim_{x\to 0}f(x)=0$.  On the other hand, for any $t\in\left[0,\frac{\pi}{2}\right]$ we have $$\frac{2}{\pi} t\leq \sin(t) \leq t \tag{1} $$
from the concavity of the sine function. By integrating over $(0,x)$, with $x\in\left[0,\frac{\pi}{2}\right]$, we get:
$$ \frac{x^2}{\pi}\leq 1-\cos(x) \leq \frac{x^2}{2}\tag{2} $$
and by integrating again:
$$ \frac{x^3}{3\pi}\leq x-\sin(x) \leq \frac{x^3}{6}\tag{3} $$
By $(2)$ and $(3)$ we have that in a right neighbourhood of the origin the ratio $\frac{x-\sin(x)}{1-\cos x}$ is bounded between $\frac{2}{3\pi}x$ and $\frac{\pi}{6}x$. It follows that $f(x)$ is (Lipschitz-)continuous at the origin and $\lim_{x\to 0}f(x)=0$.
A: Lemma: If $f$ is continuous, nonnegative, and increasing on some $[0,b],$ then $0\le \int_0^x f(t)\,dt \le x\cdot f(x)$ for $x\in [0,b].$ Proof: Obvious.
Now $x - \sin x = \int_0^x (1-\cos t)\, dt.$ Since $1-\cos t$ is continuous, nonnegative, and increasing on $[0,\pi],$ we see by the lemma that $0\le x-\sin x \le x(1-\cos x)$ for $x\in [0,\pi].$ For $x\in (0,\pi]$ we then have
$$0\le \frac{x-\sin x}{1-\cos x} \le \frac{x(1-\cos x)}{1-\cos x} = x.$$
Thus our limit from the right equals $0.$ Because we're dealing with an odd function, the limit from the left is $0$ also, showing that the limit of interest is $0.$
A: First of all you can note that as $x \to 0^{+}$ we have $$\sin x < x < \tan x$$ and therefore $$0 < x - \sin x < \tan x - \sin x = \tan x (1 - \cos x)$$ or $$0 < \frac{x - \sin x}{1 - \cos x} < \tan x$$ and applying Squeeze theorem we get $$\lim_{x \to 0^{+}}\frac{x - \sin x}{1 - \cos x} = 0$$ The case $x \to 0^{-}$ is now easily handled by putting $x = -t$ and noting that $t \to 0^{+}$.
