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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 :)

  • 1
    Use Taylor expansion.2012-10-19
  • 0
    Epsilon-Delta argument...have fun with that one.2012-10-19
  • 12
    ... without the Axiom of Choice, without using the letter 'a', without a net...2012-10-19
  • 1
    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$2012-10-19
  • 0
    Why (-1) downvoter ?2012-10-19
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    @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.2012-10-19
  • 0
    possible duplicate of [Evaluation of $\lim\limits\_{x\rightarrow0} \frac{\tan(x)-x}{x^3}$](http://math.stackexchange.com/questions/157903/evaluation-of-lim-limits-x-rightarrow0-frac-tanx-xx3)2012-10-20
  • 2
    See in particular this answer, which also covers the case with $\sin$ instead of $\tan$: http://math.stackexchange.com/a/158134/12422012-10-20

5 Answers 5

35

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 $24L = 4 \implies L = \dfrac16$

  • 3
    Thanks for ending this madness :D.2012-10-19
  • 12
    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 solve2012-10-19
  • 4
    How do you show that the $\lim$ exists in the firs place?2012-10-19
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    Yes, that is a good point. Let me think of a method to prove that limit exists and then add it to the post.2012-10-19
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    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.2012-10-19
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    @Graphth Yes. That was the only way that came to my mind to prove the existence of the limit :)2012-10-19
  • 0
    This is somewhat similar to [this question](http://math.stackexchange.com/questions/184053/how-to-find-lim-limits-x-to0-fracex-1-xx2-without-using-lhopitals-r/184245#184245) (and answer) of mine (which didn't get so much upvotes).2012-10-20
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    @enzotib Don't worry about up-votes. The voting, to the most part, is at best random on this website. So no point losing sleep over it. I have +1ed your answer :)2012-10-20
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    Oh yes, sometimes some good questions pass over heads for unpredictable reasons, having posted on Sunday or similar.2012-10-20
  • 0
    We are waiting the proof of the existence of the limit. (Without l'Hopital's rule, or the Taylor expansion, finite or infinite :-) )2012-10-20
  • 0
    See this: http://math.stackexchange.com/a/158134/12422012-10-20
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    @HansLundmark Thanks. Instead of closing this question, I would suggest to paste your proof of $\dfrac{x - \sin(x)}{x^3}$ here, since the other question asks for $\dfrac{x - \tan(x)}{x^3}$.2012-10-20
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    Well, then both the question *and* the answer would be duplicated, so I think a link is enough... ;-)2012-10-20
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    @HansLundmark Thanks for the link. In that page there are a lot of interesting proofs.2012-10-23
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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.

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    yes, yes . another method ?2012-10-19
  • 0
    This is the definition of $\sin$, so I any other method would necessarily be derived from this.2012-10-19
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    yes, I have agree with you. I forgot to mention it.2012-10-19
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    @IHaveAStupidQuestion That is **one** definition of $\sin$. There are others.2012-10-19
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    @Peter But this is the one normally done in introductory calculus courses and the most natural one.2012-10-19
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    @IHaveAStupidQuestion I was taught that $\sin$ was defined as $$\sin := \frac{\text {Opposite}}{\text{Hypotenuse}}$$ Abramowitz and Stegun define $$\sin x := \frac{e^{i x}-e^{- i x}}{2}$$ One may also define $\sin$ as a solution to a differential equation. Many possible definitions exist.2012-10-19
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    @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.2012-10-19
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    You could make up arbitrarily many nice formulas that are equal to $sin(x)$. I think this discussion is pointless.2012-10-19
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    @IHaveAStupidQuestion But clearly there exist many nice definitions for sine!2012-10-19
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    Totally agreed. But you will agree that using the Taylor series/exponential function is the most canonical and most frequently used. You said it yourself, its even in Abramowitz/Stegun.2012-10-19
  • 0
    Why did you delete the cool continued fraction? :D2012-10-19
  • 0
    @IHaveAStupidQuestion The fraction was big and somewhat unnecessary. However, I also point out that $\sin$ was known before the series was!2012-10-19
  • 0
    let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/6183/discussion-between-ihaveastupidquestion-and-argon)2012-10-19
8

Short answer: function discussion. Long answer (details) is here.

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)

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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 $$ |\sin x - x + \frac{x^3}{6} | \leq \epsilon \frac{|x|^3}{6}$$ If $0 < |x| < \delta$, then dividing through by $|x^3|$ gives: $$ | \frac{\sin x - x}{x^3} +\frac{1}{6} | \leq \frac{\epsilon}{6} < \epsilon$$ The desired limit follows.

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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*}