# 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.

• $\sin$ is odd, so $\sin'' = 0$ (or you can just use $\sin'' = -\sin$). Thus you have $\sin x = x + 0x^2 + O(x^2)$ and that's it. Apr 19, 2012 at 19:30
• @savick01 Maybe he doesn't want Taylor polynomials either.
– Pedro
Apr 19, 2012 at 19:31
• @PeterT.off Maybe, but it's not very hard. By definition of the derivative he has: $f(x) = f(0) + f'(0)x + O(x)$, so also: $f'(x) = f'(0) + f''(0)x + O(x)$ and by integrating that he gets $f(x) = f(0) + f'(x) x + f''(0) \frac{x^2}{2} + O(x^2)$, so he can even include the proof in his work. Apr 19, 2012 at 19:36
• @Peter T.off This is implcit usage of L'Hopital's rule Apr 19, 2012 at 19:36
• See groups.google.com/group/sci.math/msg/5e39a97048392a83 (sci.math thread "The approximation sin(x) = x - (1/6)x^3"), which is also at mathforum.org/kb/message.jspa?messageID=6865083 I started to LaTeX the relevant part (the part titled "NON-CALCULUS PROOF THAT sin(x) > x - (1/6)x^3"), but quickly realized that I simply don't have the time to rewrite it now. Apr 19, 2012 at 19:48

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

• Wait. That's only obvious because you know that $\sin(x)\over x$ converges to 1. So wouldn't you need to show that as well, within the constraints given in the question? Apr 19, 2012 at 20:09
• @MarkAdler: showing that $\sin(x)/x$ is 1 follows from the elementary inequality $\sin (x) < x$, $0<x<\pi/2$ (you can convince yourself of this inequality by staring at a unit circle long enough). Apr 20, 2012 at 14:46
• First off, my point is that the answer above just converts one limit problem into another, without providing the solution for the second. It still does not. Second, how does that follow? $x/2<x$ for $0<x<\pi/2$, however ${x/2\over x}$ does not go to $1$. It goes to $1/2$. Apr 20, 2012 at 20:34

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:

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

• Of course, you also need the geometric proof that $\lim_{x\to 0} \frac{\sin x}{x} = 1$. Apr 19, 2012 at 20:16

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

• I might have missed something, but doesn't this answer show only that if $\lim\frac{x-\sin(x)}{x^3}$ exists then it is equal to $1/6$? Dec 27, 2014 at 10:27

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 $$x_0$$ $$\lim_{x\to 0}(x-0)\space+\space\lim_{x\to 0}f(0)\space=\space 0\space+\space f(0)\space=\space 0\space+\space 0 \space-\space \sin 0 \space=\space 0$$