$$\lim_{x\to 0}\frac{\sin(\tan x)-\tan(\sin x)}{\arcsin(\arctan x)-\arctan(\arcsin x)}$$ Hopital's rule and Taylor expansion are practically impossible. Is there a better way to do it (mathematica gives the answer $1$). This limit is of the form $$\lim_{x\to 0}\frac{f(x)-g(x)}{g^{-1}(x)-f^{-1}(x)}$$ but first $f$ and $g$ are not monotone on any neighborhood of $0$ and second how to interpret this limit graphically.
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2$f(x)=\sin(\tan x)$ is monotonic in a neighborhood of $0$: it's the composition of two increasing functions. – egreg Oct 27 '16 at 17:15
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@egreg My fault. I had $\sin(1/x)$ in my mind. – user5402 Oct 27 '16 at 17:17
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Isn't $f(x)=\sin(\tan x)$ increasing in a neighborhood of $0$? I think it is, and also $g(x)=\tan(\sin(x))$. But, how does it help with the problem? – Darío G Oct 27 '16 at 17:18
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@Wore It garanties the existence of $f^{-1}$. – user5402 Oct 27 '16 at 17:19
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Taylor expansion is hard but not practically impossible. Note that the functions involved are all odd. After a few computations we find that $$\sin(\tan x)-\tan(\sin x)=-\frac{x^7}{30}+o(x^7).$$
Alternative approach. Let $f(x)=x+ax^n+o(x^n)$ and let $g(x)=x+bx^n+o(x^n)$ with $a\not=b$. Then $f^{-1}(x)=x-ax^n+o(x^n)$ and $g^{-1}(x)=x-bx^n+o(x^n)$. Hence $$\lim_{x\to 0}\frac{f(x)-g(x)}{g^{-1}(x)-f^{-1}(x)}= \lim_{x\to 0}\frac{(a-b)x^n+o(x^n)}{(-b+a)x^n+o(x^n)}=1.$$ But here it is not useful because $$f(x)=\sin(\tan x)=x+\frac{x^3}{6}+o(x^3)\quad\mbox{and}\quad g(x)=\tan(\sin x)=x+\frac{x^3}{6}+o(x^3).$$ However,HERE you can find a more general approach.
Robert Z
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Ok I see why $f^{-1}(x)=x-ax^n+o(x^n)$. $f(x)-g(x)$ is the vertical distance between the 2 curves. $g^{-1}(x)-f^{-1}(x)$ is the vertical distance between the symetrical curves. – user5402 Oct 27 '16 at 17:53
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