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The limit is

$$\lim_{x\rightarrow0} \frac{x-\sin_n(x)}{x^3},$$

where $\sin_n(x)$ is the $\sin(x)$ function composed with itself $n$ times:

$$\sin_n(x) = \sin(\sin(\dots \sin(x)))$$

For $n=1$ the limit is $\frac{1}{6}$, $n=2$, the limit is $\frac{1}{3}$ and so on.

Can we define a recurrent relation upon that given hypothesis? Also, how do I involve $n$ into calculation, because the final limit will depend on it?

Any suggestions on how to tackle this?

Thank you!

Alexandru Dinu
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  • On the face of it, the question looks complicated but it is almost trivial if one accepts that the answer for $n=1$ is $1/6$. See my answer below. +1 for this nice question. – Paramanand Singh Jul 01 '15 at 03:02

4 Answers4

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It's almost obvious that the limit is $n/6$. We can write $$x - \sin_{n} x = (x - \sin x) + (\sin x - \sin_{2}x) +\cdots + (\sin_{n-1}x - \sin_{n}x)\tag{1}$$ and upon division by $x^{3}$ each term tends to $1/6$.

Explanation: Since $\sin x \to 0$ as $x \to 0$, it follows (by repeatedly replacing $x$ with $\sin x$) that $\sin_{n} x \to 0$ as $x \to 0$ for all integers $n > 0$. Again we have the fundamental limit $$\lim_{x \to 0}\frac{\sin x}{x} = 1\tag{2_1}$$ Replacing $x$ by $\sin x$ repeatedly we get the following set of limit evaluations \begin{align} \lim_{x \to 0}\frac{\sin_{2}x}{\sin x} &= 1\tag{2_2}\\ \lim_{x \to 0}\frac{\sin_{3}x}{\sin_{2} x} &= 1\tag{2_3}\\ \cdots &= \cdots\notag\\ \lim_{x \to 0}\frac{\sin_{n}x}{\sin_{n - 1} x} &= 1\tag{2_n} \end{align} Multiplying all the equations above we get $$\lim_{x \to 0}\frac{\sin_{n}x}{x} = 1,\, n > 0\tag{3}$$ Using L'Hospital's Rule we know that $$\lim_{x \to 0}\frac{x - \sin x}{x^{3}} = \lim_{x \to 0}\frac{1 - \cos x}{3x^{2}} = \lim_{x \to 0}\frac{1 - \cos^{2} x}{3x^{2}(1 + \cos x)} = \frac{1}{6}\lim_{x \to 0}\frac{\sin^{2}x}{x^{2}} = \frac{1}{6}\tag{4_1}$$ Replacing $x$ with $\sin_{n - 1}x$ (also define notation $\sin_{0}x = x$) in the above equation we get $$\lim_{x \to 0}\frac{\sin_{n - 1}x - \sin_{n}x}{\sin_{n - 1}^{3}x} = \frac{1}{6}\tag{4_2}$$ and hence using $(3)$ we get $$\lim_{x \to 0}\frac{\sin_{n - 1}x - \sin_{n}x}{x^{3}} = \lim_{x \to 0}\frac{\sin_{n - 1}x - \sin_{n}x}{\sin_{n - 1}^{3}x}\cdot\frac{\sin_{n - 1}^{3}x}{x^{3}} = \frac{1}{6}\tag{4_3}$$ for all integers $n > 0$ where by definition $\sin_{0}x = x$. Hence $$\lim_{x \to 0}\frac{x - \sin_{n}x}{x^{3}} = \sum_{k = 1}^{n}\lim_{x \to 0}\frac{\sin_{k - 1}x - \sin_{k}x}{x^{3}} = \sum_{k = 1}^{n}\frac{1}{6} = \frac{n}{6}\tag{5}$$ In routine problems like these it does not make much sense to use advanced tools like Taylor series when the answer is arrived at instantly by using basic theorems of "algebra of limits". Sadly the trend on MSE is to treat Taylor and LHR as the holy grail of limits and applying them on almost any limit problem (however trivial it might be in reality). For a problem worthy of these advanced tools please see this question.

Community
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Paramanand Singh
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    Taylor series is (if set up correctly) not really more 'advanced' than the so-called basic theorems you are using, because you essentially need the geometric definition of $\sin$, which involves either unnecessary additional geometric axioms to pin down what is perimeter or area, or unintuitive definitions of arc-length based on integration. In contrast, it is possible to start from the fundamental differential equations that give rise to the elementary functions and then define canonical solutions by power series and prove the necessary results, all without integration or additional axioms. – user21820 Jul 01 '15 at 08:54
  • On the other hand, L'Hopital rule is intrinsically different from Taylor series because it fails in 'many' cases (except on problems from a textbook) due to its convoluted nature. It should be used for proving general theorems but not for concrete problems, since asymptotic expansions (including Taylor series as a tiny subset) are stronger, more intuitive and more flexible that L'Hopital. – user21820 Jul 01 '15 at 08:57
  • @user21820: Thanks for your views on both LHR and Taylor. Note however that justifying area/arclength of various smooth curves like a circle is on simpler level than dealing with differential equations and power series (at least in my opinion). And even if I accept that a power series definition is simpler for $\sin,\cos$ I must mention that conceptually Taylor series deals of infinitely differentiable functions and thats at a much higher level than algebra of limits. LHR is comparatively on a simpler level (assumes first order derivatives). Continued in next comment. – Paramanand Singh Jul 01 '15 at 09:21
  • @user21820: The problem with LHR is that most students don't know its proof as well as its conditions of applicability. BTW I do feel that the power series definitions for $\sin, \cos $ are bit tricky if we want to establish periodic properties and I believe totally unsuitable for a trigonometry course. – Paramanand Singh Jul 01 '15 at 09:25
  • Yes that's why I restrict my comment to the specific power series of the elementary functions as definitions rather than as derived theorems about them (defined through other means). These particular series are easily proven to have the desired properties. The asymptotic form of Taylor theorem with asymptotic error term is also perfect for functions with finitely many derivatives. (It is more general than the typical Taylor theorem with explicit remainder term and hence slightly weaker, but is more useful in practice.) – user21820 Jul 01 '15 at 09:35
  • Periodic properties are easy to obtain via the exponential function (also defined via power series). I am not sure what you mean by a trigonometric course. If real analysis is involved I would recommend using power series. I agree with you that L'Hopital needs to be taught rigorously, but in some sense the fact that you consider the conditions tricky suggests that perhaps it is not so simple after all? – user21820 Jul 01 '15 at 09:42
  • @user21820: I repeat periodic properties are difficult to establish if we go via power series approach. This has nothing to do with exponential series and rather deals with showing the existence of $\pi/2$ as the smallest positive number $x$ where the series $\sum (-1)^{n}x^{2n}/(2n)!$ vanishes. And then establishing sum formulas (this is where u might make use of exponential series). Trigonometric courses (at least in our country India) are supposed to mean "introduction to $\sin, \cos$" for students of age 14 years. – Paramanand Singh Jul 01 '15 at 09:46
  • The way I would do it is to define $\pi$ as twice the least positive root of $\cos$, which can be shown to exist via the exponential series as I said, simultaneously while proving periodicity of all the elementary functions. But I see that you are talking about 14-year-olds. Why do they need to learn limits of trigonometric functions (presumably) even before they understand logic and quantifiers? I should make clear that all my comments were for teaching of real analysis and not for teaching of students who do not have a sufficient foundation in logic and proofs. – user21820 Jul 01 '15 at 09:55
  • Let us continue this discussion in chat. – Paramanand Singh Jul 01 '15 at 10:07
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Hint: Taylor series suggests that $$\sin_n x=x-\frac{n}{3!}x^3 +o(x^3)$$ Try to prove it by induction.

Vim
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  • Better and more pithy than my suggestion. – Lubin Jul 01 '15 at 02:26
  • Yes, this solves it. The limit is $\frac{n}{6}$, but how did you reach to this relation? P.S.: How can Taylor, for a given function $f$, be extended to its $n$ time composition with itself, $f_n$? – Alexandru Dinu Jul 01 '15 at 02:26
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    @AlexandruDinu provided that you are familiar with the Taylor series and little o notation: for x that tends to zero we have $$\sin x=x-\frac{1}{3!}x^3 +o(x^3)$$. Note that sin x also tends to zero, so we again have $$\sin\sin x=x-\frac{1}{3!}x^3 +o(x^3)-\frac{1}{3!}(x-\frac{1}{3!}x^3 +o(x^3))^3+o((x-\frac{1}{3!}x^3 +o(x^3))^3)=x-\frac{2}{3!}+o(x^3)$$ and so forth... – Vim Jul 01 '15 at 02:31
  • @Lubin Thanks $${}{}$$ – Vim Jul 01 '15 at 02:32
  • Thank you, sir! I now fully understand it. This is great for future references. – Alexandru Dinu Jul 01 '15 at 02:32
  • @AlexandruDinu glad that I can help :) – Vim Jul 01 '15 at 02:33
  • @AlexandruDinu: By the way, what we are doing is to leave the stricter world of Taylor series for the (in some sense) more beautiful world of asymptotic expansions, which are easier to derive than Taylor series (such as you can see in this example). This bigger world includes not only the Laurent series in Lubin's answer but many other functions too. What we must be careful with when using asymptotic expansions, however, is the exact dependencies of each asymptotic term, if not it can be easy to accidentally mix up between uniform and non-uniform continuity during manipulation. – user21820 Jul 01 '15 at 09:03
  • @AlexandruDinu: One alternative to Landau notation is to carry around explicit constants. For example, $\sin(x) \in x + [1] x^3$ where "$[a]$" denotes "${ r : |r| \le |a| }$". It is then easy to manipulate such classes without worrying about accidental quantifier switch, and it has the same proving strength as using Landau notation. – user21820 Jul 01 '15 at 09:06
  • @user21820 Yes indeed. Dealing with the general functional sequence (usually over an interval on R) is much more tricky and requires much more attention than dealing with Taylor series in an infinitesimal neighbourhood. – Vim Jul 01 '15 at 09:09
  • @user21820 Do you mean the big O notation $\mathcal O$? – Vim Jul 01 '15 at 09:11
  • @Vim: Yes Landau notation refers to all of Big-O, Little-O, Omega, Little-Omega and Theta notations. =) – user21820 Jul 01 '15 at 09:11
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This is not an answer but it is too long for a comment.

The problem being really interesting, I tried to have an extra information : how is approached the limit ?

Doing almost the same as in Paramanand Singh's and Vim's answers and comments, I used one extra term for the Taylor series of $\sin(x)$ and arrive to $$sin_n(x)=x-\frac n 6 x^3 -\big(\frac n{30}-\frac{n^2}{24}\big) x^5+\cdots$$ which means that $$ \frac{x-\sin_n(x)}{x^3}=\frac n 6+\big(\frac n{30}-\frac{n^2}{24}\big) x^2+\cdots$$

Claude Leibovici
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  • +1 for the extra term calculation. BTW I must ask how do you calculate Taylor series so easily (and fast)? I always try to avoid multiplication and division of power series in order to save the pain of calculating coefficients. Its too much number crunching for me. – Paramanand Singh Jul 01 '15 at 03:46
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    @ParamanandSingh. Thank you. I must confess that I am in love with Taylor series for nearly sixty years. This case was very easy because of the clarity of the pattern. – Claude Leibovici Jul 01 '15 at 04:23
  • Lovers of power series, Unite! – Lubin Jul 01 '15 at 12:34
  • @ClaudeLeibovici +1 for Taylor series. (Btw I always try to avoid using LHR as much as I can – Vim Jul 01 '15 at 13:00
  • @Vim. I always avoid L'Hopital which can be a nightmare. However, it is good th learn it ! Cheers. – Claude Leibovici Jul 01 '15 at 14:00
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The way to tackle it is this: think of your functions as living in the ring $\Bbb R[[x]]$ and the field $\Bbb R((x))$. These are the formal power series ring and its fraction-field, the field of Laurent series with only finitely many terms with negative exponents. In the ring of power series, you may always cut off by ignoring terms of degree greater than or equal to a given $n$. This amounts to working in $\Bbb R[x]/(x^n)$. You can easily show how $x+ax^3$ and $x+bx^3$ compose together, modulo $(x^4)$. This observation will give you your answer.

Ali Caglayan
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Lubin
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