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find this follow strange limit

$$\lim_{x\to 0}\dfrac{\arcsin{\arctan{\sin{\tan{\arcsin{\arctan{x}}}}}}-\sin{\tan{\arcsin{\arctan{\sin{\tan{x}}}}}}}{\arctan{\arcsin{\tan{\sin{\arctan{\arcsin{x}}}}}}-\tan{\sin{\arctan{\arcsin{\tan{\sin{x}}}}}}}\cdots (1)$$

and I think this limit is $1$,But I can't have proof.

so I have found this follow same limit(But this is not hard)

$$\lim_{x\to0}\dfrac{\sin{\tan{x}}-\tan{\sin{x}}}{\arcsin{\arctan{x}}-\arctan{\arcsin{x}}}=1$$

and this limit is very famous,and this problem have two methods(maybe have more methods) this proof [1] can see:http://math.berkeley.edu/~giventh/lim.pdf

and the proof [2] can see http://www.stewartcalculus.com/data/ESSENTIAL%20CALCULUS%202e/upfiles/instructor/ess_wp_0807a_inst.pdf

and the proof [3] can see: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=67&t=360183

Now $(1)$ How prove it.Thank you .

math110
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  • Please describe what the "..." is referring to in $(1)$. – abiessu Sep 18 '13 at 16:46
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    Also order of operations would be nice. – Kaster Sep 18 '13 at 16:47
  • How operations ?Thank you @Kaster – math110 Sep 18 '13 at 16:49
  • I hope see this problem solution,because I have try the same method to sometimes,But I can't prove it.Thank you veryone can post this solution. – math110 Sep 18 '13 at 16:52
  • @oldrinb, my idea the same as you,But I failed at last – math110 Sep 18 '13 at 16:54
  • @Kaster let $$f=\arcsin\circ\arctan\circ\sin\circ\tan\g=\sin\circ\tan\circ\arcsin\arctan$$ with similar $h,k$. Now let $$f^\infty=\lim_{n\to\infty} f^n$$ and similarly for $g,h,k$. He wants $$\lim_{x\to0}\frac{f^\infty-g^\infty}{h^\infty-k^\infty}$$ – obataku Sep 18 '13 at 16:58
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    @oldrinb Your $f^\infty$ does not exist outside $x=0$. $f^n(x)=x-\frac n{30}x^7+O(x^9)$. I guess that the OP's "$\ldots$" just stands for "$=?$". – Hagen von Eitzen Sep 18 '13 at 17:12
  • oh, oops -- sorry friend, I was on my phone and could not see clearly. Thanks @HagenvonEitzen – obataku Sep 18 '13 at 17:13
  • @HagenvonEitzen what if we mean $f^\infty$ in terms of pointwise convergence at $x=0$? then we do still get $1$ yes? – obataku Sep 18 '13 at 17:15
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    @oldrinb If $f^\infty$ etc. are only defined at $x=0$, there is no use in taking $\lim_{x\to 0}$ and all we have is an undefined $\frac{0-0}{0-0}$. - But you are right insofar as $\lim_{x_\to 0}\frac{f^n-g^n}{h^n-k^n}=1$ for all $n$ (cf. what I said about "adding more rounds" in my solution) – Hagen von Eitzen Sep 18 '13 at 17:32
  • The Maple command $$limit((arcsin(arctan(sin(tan(arcsin(arctan(x))))))-sin(tan(arcsin(arctan(sin(tan(x)))))))/(arctan(arcsin(tan(sin(arctan(arcsin(x))))))-tan(sin(arctan(arcsin(tan(sin(x))))))), x = 0)$$ outputs $1.$ – user64494 Sep 18 '13 at 17:50

1 Answers1

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It suffices to know that $$\begin{align}\sin(x)&=x-\tfrac16x^3+O(x^4)\\ \tan(x) &= x+\tfrac13x^3+O(x^4)\\\arcsin(x)&=x+\tfrac16x^3+O(x^4)\\\arctan(x) &= x-\tfrac13x^3+O(x^4)\end{align}$$ and observe that $f(x)=x+ax^3+O(x^4)$, $g(x)=x+bx^3+O(x^4)$ implies that $f(g(x))=x+(a+b)x^3+O(x^4)$. [This also shows why the inverse functions have the coefficient of $x^3$ negated.] Especially, when working only modulo $O(x^4)$ composition becomes commutative and most of the $\sin$ and $\tan$ cancel against $\arcsin$ and $\arctan$, respectively. This way, the numerator becomes $$ x+(\tfrac16-\tfrac13)x^3 - x- (\tfrac13-\tfrac16)x^3+O(x^4)=-\tfrac13x^3+O(x^4)$$ [we only need that it is not $0+O(x^4)$ accidentally] and the denominator the same (by commutativity!), hence the quotient is $$ 1+O(x)$$ and hence the limit as $x\to 0$ is $1$. The result does not change when you add more complextity by adding additional rounds in the same pattern.

Remark: It is not necessariy for finding the limit, but more precisely the quotient is $1-\tfrac 12x^6+O(x^8)$.

In a more abstract formulation, I used this: The set of germs of functions $f$ analytic at $0$ with $f(0)=0$, $f'(0)=1$, $f''(0)=0$ form a group under composition and $f\mapsto f'''(0)$ is a group homomorphism to $\mathbb R$. A similar statement does not hold if we increase precision in order to handle more than one nontrivial coefficient.

Hagen von Eitzen
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