Solving $\lim\limits_{x\to0} \frac{x - \sin(x)}{x^2}$ without L'Hospital's Rule.

How to solve $\lim\limits_{x\to 0} \frac{x - \sin(x)}{x^2}$ Without L'Hospital's Rule? you can use trigonometric identities and inequalities, but you can't use series or more advanced stuff.

The given expression is odd; therefore it is enough to consider $x>0$. We then have $$0&lt{x-\sin x\over x^2}&lt{\tan x -\sin x\over x^2}=\tan x\ {1-\cos x\over x^2}={\tan x\over2}\ \Bigl({\sin(x/2)\over x/2}\Bigr)^2\ ,$$ and right side obviously converges to $0$ when $x\to0+$.

We will in fact prove that $\lim_{x \to 0} \dfrac{x-\sin(x)}{x^3} = \dfrac16$. This implies that $\lim_{x \to 0} \dfrac{x-\sin(x)}{x^2} = 0$.

Let $$S=\lim_{x \to 0} \dfrac{x-\sin(x)}{x^3}$$ Replacing $x$ by $2y$, we get that \begin{align} S & = \lim_{y \to 0} \dfrac{2y-\sin(2y)}{(2y)^3} = \lim_{y \to 0} \dfrac{2y-2 \sin(y) \cos(y)}{8y^3}\\ & = \lim_{y \to 0} \dfrac{2y - 2 \sin(y) + 2 \sin(y) - 2 \sin(y) \cos(y)}{8y^3}\\ & = \lim_{y \to 0} \dfrac{2 y - 2 \sin(y)}{8y^3} + \lim_{y \to 0} \dfrac{2 \sin(y) - 2 \sin(y) \cos(y)}{8y^3}\\ & = \dfrac14 \lim_{y \to 0} \dfrac{y-\sin(y)}{y^3} + \dfrac14 \lim_{y \to 0} \dfrac{\sin(y) (1 - \cos(y))}{y^3}\\ & = \dfrac{S}4 + \dfrac14 \lim_{y \to 0} \dfrac{\sin(y) 2 \sin^2(y/2)}{y^3}\\ & = \dfrac{S}4 + \dfrac18 \lim_{y \to 0} \dfrac{\sin(y)}{y} \dfrac{\sin^2(y/2)}{(y/2)^2}\\ & = \dfrac{S}4 + \dfrac18 \lim_{y \to 0} \dfrac{\sin(y)}{y} \lim_{y \to 0} \dfrac{\sin^2(y/2)}{(y/2)^2}\\ & = \dfrac{S}4 + \dfrac18\\ \dfrac{3S}4 & = \dfrac18\\ S & = \dfrac16 \end{align}

Hence, $$\lim_{x \to 0} \dfrac{x-\sin(x)}{x^2} = \lim_{x \to 0} \left(\dfrac{x-\sin(x)}{x^3} \right)x = \left(\lim_{x \to 0} \dfrac{x-\sin(x)}{x^3} \right) \left( \lim_{x \to 0} x \right) = \dfrac{\lim_{x \to 0} x}6 = 0$$

This can be done geometrically.

Surprisingly, two answers I wrote in this regard(geometric proofs of limits) before can be combined to give a solution for this.

$$\lim_{x \to 0} \frac{ \tan x - x}{x^2} = 0 \tag{1}$$

A geometric proof of that can be found here: Limit, solution in unusual way

$$\lim_{x \to 0} \frac{1 - \cos x}{x} = 0 \tag{2}$$

A geometric proof of that can be found here: Finding the limit of $(1-\cos(x))/x$ as $x\to 0$ with squeeze theorem

To combine the two:

$$\tan x - x = \frac{\sin x - x \cos x}{\cos x} = \frac{(\sin x - x) + x(1 - \cos x)}{\cos x}$$