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.
 A: 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:
$$\tan x - x = \frac{\sin x - x \cos x}{\cos x} = \frac{(\sin x  - x) + x(1 - \cos x)}{\cos x}$$
A: 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, $$\lim_{x \to 0} \dfrac{x-\sin(x)}{x^2} = \lim_{x \to 0} \left(\dfrac{x-\sin(x)}{x^3} \right)x = \left(\lim_{x \to 0} \dfrac{x-\sin(x)}{x^3} \right) \left( \lim_{x \to 0} x \right) = \dfrac{\lim_{x \to 0} x}6 = 0$$
A: 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 $x\to0+$.
A: 1.Divide both nominator and denominator by $(1-\cos x)$ and apply product rule for limits $$\lim_{x\to 0}\Biggl(\frac{\frac{x-\sin x}{1-\cos x}}{\frac{x^2}{1-\cos x}}\Biggl)\space=\space\lim_{x\to 0}\Biggl(\frac{x-\sin x}{1-\cos x}\Biggl)\space \lim_{x\to 0}\Biggl(\frac{1-\cos x}{x^2}\Biggl)$$ 2. Notice the standard limit in this expression and use it in the next step $$\lim_{x\to 0}\frac{1-\cos x}{x^2}\space=\space\frac12$$ 3.$$\frac12\lim_{x\to 0}\Biggl(\frac{x-\sin x}{1-\cos x}\Biggl)$$ 4. Notice that denominator is derivative of nominator. We can modify the equation a little bit by $$f(x)\space=\space x-\sin x$$ $$f'(x)\space=\space 1-\cos x$$ 5. If we substitute $x_0$ for $0$ and accept $\frac{f(x)}{f'(x)}$ is continuous at $x_0$ , we get $$\frac12\lim_{x\to x_0}\frac{f(x)}{f'(x)}\space=\space \frac{f(x_0)}{f'(x_0)}\space=\space (x-x_0)\space+\space f(x_0)$$  7. Finally substitute $0$ for $x_0$ $$\lim_{x\to 0}(x-0)\space+\space\lim_{x\to 0}f(0)\space=\space 0\space+\space f(0)\space=\space 0\space+\space 0 \space-\space \sin 0 \space=\space 0$$
