# Determine $\lim_{x \to 0}{\frac{x-\sin{x}}{x^3}}=\frac{1}{6}$, without L'Hospital or Taylor

How can I prove that $$\lim_{x \to 0}{\frac{x-\sin{x}}{x^3}}=\frac{1}{6}$$

without using L'Hospital or Taylor series?

thanks :)

• Use Taylor expansion. – J.R. Oct 19 '12 at 20:20
• ... without the Axiom of Choice, without using the letter 'a', without a net... – rschwieb Oct 19 '12 at 20:41
• GH Hardy "Pure Mathematics" Examples XLVI (page 237 in Tenth Edition) has some nice exercises starting with $x-\sin x$ and $\tan x-x$ are increasing for suitable small positive $x$ – Mark Bennet Oct 19 '12 at 20:42
• @Iuli Perhaps because the question was changed to exclude a possibility after an answer was given - which was, I think, intended to invite alternative answers, but misguided (rude) as a way of doing so. The person who put the answer has had to amend it to explain why it was given. Instead you might have put - "there is a nice Taylor Series answer already, but I'm also looking for other ideas and ways of attacking this problem". I was not the down voter. – Mark Bennet Oct 19 '12 at 21:21
• See in particular this answer, which also covers the case with $\sin$ instead of $\tan$: math.stackexchange.com/a/158134/1242 – Hans Lundmark Oct 20 '12 at 13:54

Let $L = \lim_{x \to 0} \dfrac{x - \sin(x)}{x^3}$. We then have \begin{align} L & = \underbrace{\lim_{y \to 0} \dfrac{3y - \sin(3y)}{27y^3} = \lim_{y \to 0} \dfrac{3y - 3\sin(y) + 4 \sin^3(y)}{27y^3}}_{\sin(3y) = 3 \sin(y) - 4 \sin^3(y)}\\ & = \lim_{y \to 0} \dfrac{3y - 3\sin(y)}{27 y^3} + \dfrac4{27} \lim_{y \to 0} \dfrac{\sin^3(y)}{y^3} = \dfrac{3}{27} L + \dfrac4{27} \end{align} This gives us $24L = 4 \implies L = \dfrac16$

• Thanks for ending this madness :D. – J.R. Oct 19 '12 at 20:49
• Dont you need first to prove that the limit exists? For example $\lim_ {x \rightarrow \infty} \cos (x)=\lim_ {x \rightarrow \infty} \cos (2x)=\lim_ {x \rightarrow \infty} \cos (x) ^2-1\Rightarrow a=2a^2-1$ and solve – clark Oct 19 '12 at 20:53
• How do you show that the $\lim$ exists in the firs place? – Hagen von Eitzen Oct 19 '12 at 20:53
• Yes, that is a good point. Let me think of a method to prove that limit exists and then add it to the post. – user17762 Oct 19 '12 at 20:55
• It exists by l'Hopital's rule, or the Taylor expansion. :) Seriously, though, assuming you can prove the existing somewhere, that is a really great idea I would never have thought of. – GeoffDS Oct 19 '12 at 21:22

Note: The original question didn't say anything about not using Taylor series. Then, after I answered this, the OP changed the question and said didn't want Taylor series either.

Another option would be using power series (aka Maclaurin series aka Taylor series centered at 0)

$$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots$$

At this point, it's pretty easy.

• yes, yes . another method ? – Iuli Oct 19 '12 at 20:21
• This is the definition of $\sin$, so I any other method would necessarily be derived from this. – J.R. Oct 19 '12 at 20:24
• @IHaveAStupidQuestion That is one definition of $\sin$. There are others. – Pedro Tamaroff Oct 19 '12 at 20:26
• @Argon The first definition is the one used in high school. The second definition is literally the same as the Taylor expansion by using the definition of exp. – J.R. Oct 19 '12 at 20:33
• You could make up arbitrarily many nice formulas that are equal to $sin(x)$. I think this discussion is pointless. – J.R. Oct 19 '12 at 20:35

First step. Your fraction is an even function so it is enough to consider the right limit.

Second step. Define the functions $f(x):=\sin(x)-x+\frac{1}{6}x^3$ and $g(x):=\sin(x)-x+\frac{1}{6}x^3-\frac{1}{120}x^5$ on the interval $[0, 0.1]$.

Third step. Prove that $f$ is strictly increasing and $g$ is strictly decreasing. ($f(0)=0$, it is enough $f'>0$. $f'(0)=0$, so it is enough $f''>0$ and so on. Similarly for $g$.

Forth step. From the third step $$x-\frac{1}{6}x^3<\sin(x)<x-\frac{1}{6}x^3+\frac{1}{120}x^5.$$ The limit follows.

Assuming $$f$$ is sufficiently smooth, repeated application of the fundamental theorem of the calculus gives (finite Taylor expansion) $$f(x) = f(0)+f'(0)x+\frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \int_0^x (x-t)^2 \frac{(f'''(t)-f'''(0))}{2!}\, dt$$

Using the fact that $$\sin' = \cos, \cos' = -\sin$$, we can then expand $$f(x) = \sin x$$ as $$\sin x = x -\frac{x^3}{6} + \int_0^x (x-t)^2 \frac{(-\cos t +1)}{2!}\, dt$$ Let $$\epsilon>0$$, and choose $$\delta>0$$ such that if $$|t|< \delta$$, then $$|1-\cos t| < \epsilon$$. Then, replacing $$1-\cos t$$ by $$\epsilon$$ and integrating, we have the estimate $$\left|\sin x - x + \frac{x^3}{6} \right| \leq \epsilon \frac{|x|^3}{6}$$ If $$0 < |x| < \delta$$, then dividing through by $$|x^3|$$ gives: $$\left| \frac{\sin x - x}{x^3} +\frac{1}{6} \right| \leq \frac{\epsilon}{6} < \epsilon$$ The desired limit follows.

The following argument is based on the suggestions by vesszabo. (I do not restrict the functions to the interval $[0, \, 0.1]$ as vesszabo did. That would erroneous because we are looking for the limit at $0$.) It is a rigorous argument and it does avoid using Taylor's Theorem ... but it is still hokey. The choice to use the functions $f$ and $g$ is guided by Taylor's Theorem.

Demonstration

$f$ is a function defined by \begin{equation*} f(x) = \sin{x} - x + \frac{1}{6} \, x^{3} . \end{equation*} This function is differentiable, and \begin{equation*} f^{\prime}(x) = \cos{x} - 1 + \frac{1}{2} \, x^{2} . \end{equation*} Likewise, this derivative is differentiable, and \begin{equation*} f^{\prime\prime}(x) = -\sin{x} + x . \end{equation*} $f(0) = f^{\prime}(0) = 0$, but since $f^{\prime\prime}(x) > 0$ for every positive, real number $x$, $f^{\prime}(x)$ is an increasing function, and since $f^{\prime}(0) = 0$, $f^{\prime}(x) > 0$ for every positive, real number $x$. According to a corollary to the Mean Value Theorem, $f$ is an increasing function. $f(0) = 0$, and so $f(x) > 0$ for every positive, real number $x$.

$g$ is another twice-differentiable function defined by \begin{equation*} g(x) = \sin{x} - x + \frac{1}{6} \, x^{3} - \frac{1}{120} x^{5} = f(x) - \frac{1}{120} x^{5} . \end{equation*} For every positive, real number $x$, \begin{equation*} g^{\prime\prime}(x) = f^{\prime\prime}(x) - \frac{1}{6} x^{3} = - f(x) < 0 . \end{equation*} According to a corollary to the Mean Value Theorem, $g^{\prime}$ is a decreasing function. $g^{\prime}(0) = 0$, and so $g^{\prime}(x) < 0$ for every positive, real number $x$. According to a corollary to the Mean Value Theorem, $g$ is a decreasing function. $g(0) = 0$, and so $g(x) < 0$ for every positive, real number $x$.

So, for every positive, real number $x$, \begin{equation*} \frac{1}{6} \, x^{3} - \frac{1}{120} x^{5} < x - \sin{x} < \frac{1}{6} \, x^{3} , \end{equation*} and \begin{equation*} \frac{1}{6} - \frac{1}{120} x^{2} < \frac{x - \sin{x}}{x^{3}} < \frac{1}{6} . \end{equation*} According to the Squeeze Theorem, \begin{equation*} \lim_{x\to0^{+}} \frac{x - \sin{x}}{x^{3}} = \frac{1}{6} . \end{equation*}

$f$ and $g$ are odd functions. So, for every negative, real number $x$, \begin{equation*} \frac{1}{6} \, x^{3} < x - \sin{x} < \frac{1}{6} \, x^{3} - \frac{1}{120} x^{5} , \end{equation*} and \begin{equation*} \frac{1}{6} - \frac{1}{120} x^{2} < \frac{x - \sin{x}}{x^{3}} < \frac{1}{6} . \end{equation*} Again, according to the Squeeze Theorem, \begin{equation*} \lim_{x\to0^{-}} \frac{x - \sin{x}}{x^{3}} = \frac{1}{6} . \end{equation*} A function that has the same left-sided and right-sided limits at $0$ has a limit at $0$. \begin{equation*} \lim_{x\to0} \frac{x - \sin{x}}{x^{3}} = \frac{1}{6} . \end{equation*}

Reproduced from this answer, in which this answer is cited to show that $$\lim\limits_{x\to0}\frac{\sin(x)}x=1$$ and that $$0\le x\lt\frac\pi2\implies0\le\sin(x)\le x\le\tan(x)$$.

Pre-Calculus Proof that $$\boldsymbol{\lim\limits_{x\to0}\frac{x-\sin(x)}{x^3}=\frac16}$$

Assume that $$0\lt x\le\frac\pi3$$. Then, $$\cos(x)\ge\frac12$$ and $$0\le\sin(x)\le x\le\tan(x)$$. Therefore, \begin{align} \frac{x-\sin(x)}{x^3} &\le\frac{\tan(x)-\sin(x)}{x^3}\tag{1a}\\ &=\frac{\tan(x)}{x}\frac{1-\cos(x)}{x^2}\tag{1b}\\ &=\frac1{\cos(x)}\frac{\sin(x)}{x}\frac{2\sin^2(x/2)}{4\,(x/2)^2}\tag{1c}\\[6pt] &\le1\tag{1d} \end{align} Furthermore, \begin{align} &\frac{x-\sin(x)}{x^3}-\frac14\frac{x/2-\sin(x/2)}{(x/2)^3}\tag{2a}\\ &=\frac{2(x/2)-2\sin(x/2)\cos(x/2)}{8(x/2)^3}-\frac{2(x/2)-2\sin(x/2)}{8(x/2)^3}\tag{2b}\\ &=\frac{2\sin(x/2)(1-\cos(x/2))}{8(x/2)^3}\tag{2c}\\ &=\frac{2\sin(x/2)\,2\sin^2(x/4)}{8(x/2)^3}\tag{2d} \end{align} Since $$\lim\limits_{x\to0}\frac{\sin(x)}x=1$$, $$(2)$$ shows that $$\lim_{x\to0}\left(\frac{x-\sin(x)}{x^3}-\frac14\frac{x/2-\sin(x/2)}{(x/2)^3}\right)=\frac18\tag3$$ For any $$n$$, adding $$\frac1{4^k}$$ times $$(3)$$ with $$x\mapsto x/2^k$$ for $$k$$ from $$0$$ to $$n-1$$ gives \begin{align} \lim_{x\to0}\left(\frac{x-\sin(x)}{x^3}-\frac1{4^n}\frac{x/2^n-\sin\left(x/2^n\right)}{\left(x/2^n\right)^3}\right) &=\frac18\frac{1-(1/4)^n}{1-1/4}\tag{4a}\\ &=\frac16-\frac16\frac1{4^n}\tag{4b} \end{align} Thus, for any $$\epsilon\gt0$$, choose $$n$$ large enough so that $$\frac1{4^n}\le\frac\epsilon2$$. Then, $$(4)$$ says that we can choose a $$\delta\gt0$$ so that if $$0\lt x\le\delta$$, $$\frac{x-\sin(x)}{x^3}-\overbrace{\frac1{4^n}\frac{x/2^n-\sin\left(x/2^n\right)}{\left(x/2^n\right)^3}}^{\frac12[0,\epsilon]} =\frac16-\!\overbrace{\ \ \ \frac16\frac1{4^n}\ \ \ }^{\frac1{12}[0,\epsilon]}\!+\frac12[-\epsilon,\epsilon]\tag5$$ where $$[a,b]$$ represents a number between $$a$$ and $$b$$. The bounds above the braces follow from $$(1)$$ and the choice of $$n$$.

Equation $$(5)$$ says that for $$0\lt x\le\delta$$, $$\frac{x-\sin(x)}{x^3}=\frac16+[-\epsilon,\epsilon]\tag6$$ Since $$\frac{x-\sin(x)}{x^3}$$ is even, we can say that $$(6)$$ is true for $$0\lt|x|\le\delta$$, which means that $$\lim_{x\to0}\frac{x-\sin(x)}{x^3}=\frac16\tag7$$