# 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, 2012 at 20:20
• ... without the Axiom of Choice, without using the letter 'a', without a net... Oct 19, 2012 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 $x$ Oct 19, 2012 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. Oct 19, 2012 at 21:21
• See in particular this answer, which also covers the case with $\sin$ instead of $\tan$: math.stackexchange.com/a/158134/1242 Oct 20, 2012 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, 2012 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 Oct 19, 2012 at 20:53
• How do you show that the $\lim$ exists in the firs place? Oct 19, 2012 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, 2012 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. Oct 19, 2012 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, 2012 at 20:21
• This is the definition of $\sin$, so I any other method would necessarily be derived from this.
– J.R.
Oct 19, 2012 at 20:24
• @IHaveAStupidQuestion That is one definition of $\sin$. There are others.
– Pedro
Oct 19, 2012 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, 2012 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, 2012 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.

First, an identity. For $$x\ne0$$, \begin{align} \frac{2\sin(x)-\sin(2x)}{x^3} &=\frac{2\sin(x)-2\sin(x)\cos(x)}{x^3}\tag{1a}\\ &=2\frac{\sin(x)}x\frac{1-\cos(x)}{x^2}\tag{1b}\\ &=\frac2{1+\cos(x)}\frac{\sin^3(x)}{x^3}\tag{1c} \end{align} Explanation:
$$\text{(1a)}$$: $$\sin(2x)=2\sin(x)\cos(x)$$
$$\text{(1b)}$$: algebra
$$\text{(1c)}$$: $$1-\cos(x)=\frac{\sin^2(x)}{1+\cos(x)}$$

In this answer, it is shown geometrically that $$\lim\limits_{x\to0}\frac{\sin(x)}x=1$$. Thus, given $$\epsilon\gt0$$, we can choose $$\delta\gt0$$ so that if $$0\lt|x|\le\delta$$, then $$\frac2{1+\cos(x)}\frac{\sin^3(x)}{x^3}=[1-\epsilon,1+\epsilon]_\#\tag2$$ where $$[a,b]_\#$$ is a real number in the interval $$[a,b]$$.

For $$0\lt|x|\le\delta$$, apply $$(1)$$ and $$(2)$$ to $$x/2^{k+1}$$ and divide by $$2^{2k+3}$$: $$\frac{2^{k+1}\sin\left(x/2^{k+1}\right)-2^k\sin\left(x/2^k\right)}{x^3}=\frac{[1-\epsilon,1+\epsilon]_\#}{2^{2k+3}}\tag3$$ $$\lim\limits_{x\to0}\frac{\sin(x)}x=1$$ implies $$\lim\limits_{k\to\infty}2^k\sin\left(x/2^k\right)=x$$. Thus, if we sum $$(3)$$ in $$k$$, we get that for $$0\lt|x|\le\delta$$, $$\frac{x-\sin(x)}{x^3}=\frac{[1-\epsilon,1+\epsilon]_\#}6\tag4$$ Therefore, the Squeeze Theorem yields $$\lim_{x\to0}\frac{x-\sin(x)}{x^3}=\frac16\tag5$$

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*}