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Calculate the following limit: $$\lim_{x \rightarrow 0} \frac{x-\overbrace{\sin (\sin (...(\sin x)...))}^{150\ \text{times}\ \sin}}{x^3}$$

I tried applying L'Hospital's rule, but it got too messy.

Thank you in advance!

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  • $\begingroup$ Have you tried Taylor expansion? $\endgroup$ – vap Oct 15 '16 at 11:11
  • $\begingroup$ I haven't studied Taylor expansions yet. $\endgroup$ – George R. Oct 15 '16 at 11:12
  • $\begingroup$ How about the power series for $\sin x$? $\endgroup$ – hmakholm left over Monica Oct 15 '16 at 11:18
  • $\begingroup$ @HenningMakholm That's not quite different from the (previously suggested) Taylor. $\endgroup$ – yo' Oct 15 '16 at 11:19
  • $\begingroup$ @yo': True -- but it might be that the OP knew the concept under one name but not the other. $\endgroup$ – hmakholm left over Monica Oct 15 '16 at 11:24
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Let us denote

$$\phi(x;n):=\sin^n(x)=\sin(\sin(\dots\sin(x)\dots)$$

and

$$\Phi(n):=\lim_{x\to0}\frac{x-\phi(x;n)}{x^3}.$$

Then you want to compute $\Phi(150)$. First of all, check by L'Hôpital rule that

$$\lim_{x\to 0}\frac{\sin(x)}{x}=1,\quad\text{and}\quad \lim_{x\to 0}\frac{x-\sin(x)}{x^3}=\frac16.$$

Note that the first limit implies that

$$\lim_{x\to0}\frac{\sin(\sin(x))}{x}= \lim_{x\to0}\frac{\sin(\sin(x))}{\sin(x)}\frac{\sin(x)}{x}= \lim_{x\to0}\frac{\sin(\sin(x))}{\sin(x)} \lim_{x\to0}\frac{\sin(x)}{x}=1$$

and in general (by an inductive argument),

$$\lim_{x\to0}\frac{\phi(x;n)}x=1.$$

Now, write

$$\begin{align*} \Phi(n)&=\lim_{x\to0}\frac{x-\phi(x;n-1)+\phi(x;n-1)-\phi(x;n)}{x^3}\\ &=\Phi(n-1)+\lim_{x\to0}\frac{\phi(x;n-1)-\phi(x;n)}{x^3}\\ &=\Phi(n-1)+\lim_{x\to0}\frac{\phi(x;n-1)-\sin(\phi(x;n-1))}{\phi(x;n-1)^3}\left(\frac{\phi(x;n-1)}{x}\right)^3\\ &=\Phi(n-1)+\lim_{x\to0}\frac{\phi(x;n-1)-\sin(\phi(x;n-1))}{\phi(x;n-1)^3}\lim_{x\to0}\left(\frac{\phi(x;n-1)}{x}\right)^3\\ &=\Phi(n-1)+\lim_{x\to0}\frac{\phi(x;n-1)-\sin(\phi(x;n-1))}{\phi(x;n-1)^3}\\ &=\Phi(n-1)+\frac16. \end{align*}$$

This is, $\Phi$ satisfies the recursive relation

$$\Phi(n)-\Phi(n-1)=\frac16.$$

Telescoping we see that

$$\Phi(N)-\Phi(1)=\sum_{n=2}^N(\Phi(n)-\Phi(n-1))=\sum_{n=2}^N\frac16,$$

or

$$\Phi(N)=\Phi(1)+(N-1)\frac16=\frac16+(N-1)\frac16=\frac{N}6.$$

Now set $N=150$ to obtain $\Phi(150)=25$.

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Use short form of series expansion: $$ \sin(x)=x-\frac{x^3}{6}+O(x^5)\\ \sin(\sin(x))=\sin(x)-\frac{\sin^3(x)}{6}+O(\sin^5(x))\to\\ \sin(\sin(x))=x-\frac{x^3}{6}+O(x^5)-\frac{(x-\frac{x^3}{6}+O(x^5))^3}{6}+O(x^5)\to\\ \sin(\sin(x))=x-\frac{x^3}{3}+O(x^5) $$ Important part here is to notice that each additional $\sin(...)$ leads to the expression of the same form, while adding $-\frac{x^3}{6}$. Then, $$\overbrace{\sin (\sin (...(\sin x)...))}^{150\ \text{times}\ \sin}=x-150\frac{x^3}{6}+O(x^5)$$ Hence, $$\lim_{x \rightarrow 0} \frac{x-\overbrace{\sin (\sin (...(\sin x)...))}^{150\ \text{times}\ \sin}}{x^3}=\lim_{x \rightarrow 0} \frac{x-x+150\frac{x^3}{6}+O(x^5)}{x^3}=\lim_{x \rightarrow 0}(25+O(x^2))=25$$

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  • $\begingroup$ I do not see why you had a downvote ! It is well done. $+1$ $\endgroup$ – Claude Leibovici Oct 15 '16 at 11:28
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    $\begingroup$ The OP doesn't want to use Taylor expansion $\endgroup$ – user261263 Oct 15 '16 at 11:29
  • $\begingroup$ Yes, the OP doesn't want to use Taylor expansions...yet he should say so on the question itself, not in the comments. BTW, any other way seems like is going to be hell... $\endgroup$ – DonAntonio Oct 15 '16 at 11:33
  • $\begingroup$ @DonAntonio It is hell (working on it just now :-D) $\endgroup$ – yo' Oct 15 '16 at 11:36
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    $\begingroup$ @yo' I know. I just began and gave up. No time for this kind of exercises which seem pointless...unless there's some intellectual gain hidden there. Besides this, my development threatened to give $\;\frac13\;$ as a result, which seems to be wrong, so... $\endgroup$ – DonAntonio Oct 15 '16 at 11:37
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The derivative is not at all messy.

Let's denote $N=150$, $S_k(x)$ the $k$-fold sine and $C_k(x)=\cos S_{k-1}(x)$. Note that we have $S_k(0)=0$ and $C_k(0)=1$.

In this notation, the numerator is $x-S_N(x)$. Its derivative is $(x-S_N(x))'=1-C_1(x)\dotsm C_N(x)$. The derivative of this (i.e., the 2nd derivative) is $$(x-S_N(x))''=\sum_{k=1}^{N} C_1^2(x)\dotsm C_{k-1}^2(x) S_k(x) C_{k+1}(x)\dotsm C_N(x),$$ because $C_k(x)' = -C_1(x)\dotsm C_{k-1}(x) S_k(x)$ and we simply use the product rule. Therefore, the 2nd derivative of the numerator is a sum of $N$ products, each containing one sine and then some cosines.

The second derivative of the denominator is $(x^3)''=6x$. If we prove that $\lim_{x\to0}\frac{S_k(x)}{x}=1$ for all $k$, we receive (applying l'Hospital twice) the result $N/6=25$, which is correct. But this is true as we have $S_k(x)'=C_1(x)\dotsm C_k(x)$ whence $\lim_{x\to0}\frac{S_k(x)}{x}=\lim_{x\to0} \frac{C_1(x)\dotsm C_k(x)}{1}=1$.

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  • $\begingroup$ Is ""not at all messy"" ? For someone who hasn't yet even studied Taylor expansions and etc., I'd say it is very, very messy...in fact, almost unbelievably messy. Yet this is what I, with all my experience, was beginning to get. Though I must say: your proof is slick and nice, but your first line may cause the OP, and others as well, feel bad...:) +1 $\endgroup$ – DonAntonio Oct 15 '16 at 12:13
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Here is a more or less elementary calculation of the limit.

A quite commonly known limit is $$ \lim_{x\to 0}\frac{\sin x}{x}=\lim_{x\to 0}\frac{\sin x-\sin 0 }{x-0}=\sin'(0)=\cos(0)=1. \tag{1}\label{1} $$ Also, you may prove with L'Hopital that $$ \lim_{x\to 0}\frac{x-\sin x}{x^3}=\lim_{x\to 0}\frac{1-\cos x}{3x^2}=\lim_{x\to 0}\frac{\sin x}{6x}=\frac{1}{6}\lim_{x\to 0}\frac{\sin x}{x}\stackrel{\eqref{1}}=\frac{1}{6}\cdot 1=\frac{1}{6}. \tag{2}\label{2} $$ Using this, we see $$ \lim_{x\to 0}\frac{\sin^{(n)}x-\sin^{(n+1)} x}{x^3}\\ =\lim_{x\to 0}\left[\frac{\sin^{(n)}x-\sin^{(n+1)} x}{\left(\sin^{(n)}x\right)^3}\cdot\left(\frac{\sin^{(n)}x}{\sin^{(n-1)}x}\right)^3\cdot\left(\frac{\sin^{(n-1)}x}{\sin^{(n-2)}x}\right)^3\cdots\left(\frac{\sin^{(1)}x}{\sin^{(0)}x}\right)^3\right]\\ =\left(\lim_{x\to 0}\frac{\sin^{(n)}x-\sin^{(n+1)} x}{\left(\sin^{(n)}x\right)^3}\right)\cdot\left(\lim_{x\to 0}\frac{\sin^{(n)}x}{\sin^{(n-1)}x}\right)^3\cdot\left(\lim_{x\to 0}\frac{\sin^{(n-1)}x}{\sin^{(n-2)}x}\right)^3\cdots\left(\lim_{x\to 0}\frac{\sin^{(1)}x}{\sin^{(0)}x}\right)^3\\ =\left(\lim_{x\to 0}\frac{x-\sin x}{x^3}\right)\cdot\left(\lim_{x\to 0}\frac{\sin x}{x}\right)^3\cdot\left(\lim_{x\to 0}\frac{\sin x}{x}\right)^3\cdots\left(\lim_{x\to 0}\frac{\sin x}{x}\right)^3\stackrel{\eqref{1}\&\eqref{2}}=\frac{1}{6}. \tag{3}\label{3} $$ Where $\sin^{(n)}x:=\overbrace{\sin(\sin(...\sin(x)...))}^{n\text{ times}}$ and $\sin^{(0)}x:=x$. We used the fact that if for two functions $f,g$ we have $$ \lim_{x\substack{\to\\ \neq}a}g(x)=b\qquad\text{and}\qquad \lim_{x\substack{\to\\ \neq}b}f(x)=l $$ and $g(x)\neq b$ in a neighborhood of $a$ then $$ \lim_{x\substack{\to\\ \neq}a}f(g(x))=l. $$

Finally we conclude $$ \lim_{x\to 0}\frac{x-\sin^{(n)} x}{x^3}=\lim_{x\to 0}\left[\frac{x-\sin^{(1)}x+\sin^{(1)}x-\sin^{(2)}x+...+\sin^{(n-1)}-\sin^{(n)} x}{x^3}\right]\\ =\left(\lim_{x\to 0}\frac{x-\sin^{(1)}x}{x^3}\right)+\left(\lim_{x\to 0}\frac{\sin^{(1)}x-\sin^{(2)}x}{x^3}\right)+...+\left(\lim_{x\to 0}\frac{\sin^{(n-1)}-\sin^{(n)} x}{x^3}\right)\\ =\overbrace{\frac{1}{6}+...+\frac{1}{6}}^{n\text{ times}}\stackrel{\eqref{3}}=\frac{n}{6} \tag{4}\label{4} $$ and therefore $$ \lim_{x\to 0}\frac{x-\sin^{(150)} x}{x^3}\stackrel{\eqref{4}}=\frac{150}{6}=25. $$

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