Crafty solutions to the following limit The following problem came up at dinner, I know some ways to solve it but they are quite ugly and as some wise man said: There is no place in the world for ugly mathematics.
These methods are using l'Hôpital, but that becomes quite hideous very quickly or by using series expansions.
So I'm looking for slick solutions to the following problem:

Compute $\displaystyle \lim_{x \to 0} \frac{\sin(\tan x) - \tan(\sin x)}{\arcsin(\arctan x) - \arctan(\arcsin x)}$.

I'm curious what you guys will make of this.
 A: Some details get lost because of the fact that $\displaystyle\lim_{x\to0}\frac{f(x)}{x}=1$. I have adjusted this answer to accept $\displaystyle\lim_{x\to0}\frac{f(x)}{x}=a$ to expose those details.

If $\displaystyle\lim_{x\to0}\frac{f(x)}{x}=a$, then $f'(f^{-1}(0))=f'(0)=a$. Furthermore, if $f^{(k)}(0)=0$ for $1<k<n$, but $f^{(n)}(0)\not=0$, we can use L'Hopital twice to get
$$
\begin{align}
\lim_{x\to0}\frac{f(x)-ax}{x/a-f^{-1}(x)}
&=\lim_{x\to0}\frac{f'(x)-a}{1/a-1/f'(f^{-1}(x))}\\
&=af'(f^{-1}(0))\;\lim_{x\to0}\frac{f'(x)-a}{f'(f^{-1}(x))-a}\\
&=a^2\cdot\lim_{x\to0}\frac{f''(x)}{f''(f^{-1}(x))/f'(f^{-1}(x))}\\
&=a^2f'(f^{-1}(0))\;\lim_{x\to0}\frac{f''(x)}{f''(f^{-1}(x))}\\
&=a^3\cdot\lim_{x\to0}\frac{f''(x)}{f''(f^{-1}(x))}\tag{1}
\end{align}
$$
Note that if $f^{(k)}(0)=0$, L'Hopital yields
$$
\begin{align}
a^{k+1}\cdot\lim_{x\to0}\frac{f^{(k)}(x)}{f^{(k)}(f^{-1}(x))}
&=a^{k+1}\cdot\lim_{x\to0}\frac{f^{(k+1)}(x)}{f^{(k+1)}(f^{-1}(x))/f'(f^{-1}(x))}\\
&=a^{k+1}f'(f^{-1}(0))\;\lim_{x\to0}\frac{f^{(k+1)}(x)}{f^{(k+1)}(f^{-1}(x))}\\
&=a^{k+2}\cdot\lim_{x\to0}\frac{f^{(k+1)}(x)}{f^{(k+1)}(f^{-1}(x))}\tag{2}
\end{align}
$$
Using $(1)$ and repeating $(2)$, we get that
$$
\lim_{x\to0}\frac{f(x)-ax}{x/a-f^{-1}(x)}=a^{n+1}\frac{f^{(n)}(0)}{f^{(n)}(f^{-1}(0))}=a^{n+1}\tag{3}
$$
Suppose $g^{(k)}(0)=0$ for $0\le k<m$ and $g^{(m)}(0)\not=0$ and $\displaystyle\lim_{x\to0}\frac{h(x)}{x}=1$. Then
$$
\begin{align}
\lim_{x\to0}\frac{g^{(k)}(h(x))}{g^{(k)}(x)}
&=\lim_{x\to0}\frac{g^{(k+1)}(h(x))h'(x)}{g^{(k+1)}(x)}\\
&=h'(0)\lim_{x\to0}\frac{g^{(k+1)}(h(x))}{g^{(k+1)}(x)}\\
&=\lim_{x\to0}\frac{g^{(k+1)}(h(x))}{g^{(k+1)}(x)}\\
&=\frac{g^{(m)}(h(0))}{g^{(m)}(0)}\\
&=1\tag{4}
\end{align}
$$

To finish things off, let $f(x)=\sin(\tan(\arcsin(\arctan(x))))$, $h(x)=\tan(\sin(x))$, and $a=1$. Then,
$$
\begin{align}
&\lim_{x\to0}\frac{\sin(\tan(x))-\tan(\sin(x))}{\arcsin(\arctan(x))-\arctan(\arcsin(x))}\\
&=\lim_{x\to0}\frac{f(h(x))-ah(x)}{h^{-1}(x/a)-h^{-1}(f^{-1}(x))}\\
&=\lim_{x\to0}\frac{f(h(x))-ah(x)}{h(x)/a-f^{-1}(h(x))}\frac{h(x)/a-f^{-1}(h(x))}{x/a-f^{-1}(x)}\frac{x/a-f^{-1}(x)}{h^{-1}(x/a)-h^{-1}(f^{-1}(x))}\\
&=a^{n+1}\tag{5}
\end{align}
$$
because
$$
\lim_{x\to0}\frac{f(h(x))-ah(x)}{h(x)/a-f^{-1}(h(x))}=a^{n+1}
$$
by $(3)$, and
$$
\lim_{x\to0}\frac{h(x)/a-f^{-1}(h(x))}{x/a-f^{-1}(x)}=1
$$
by $(4)$ using $g(x)=x/a-f^{-1}(x)$, and
$$
\lim_{x\to0}\frac{x/a-f^{-1}(x)}{h^{-1}(x/a)-h^{-1}(f^{-1}(x))}=\frac{1}{1/h'(h^{-1}(0))}=1
$$

Summary: Letting $g(x)=ah(x)$, we get that if
$$
\lim_{x\to0}\frac{f(x)}{x}=\lim_{x\to0}\frac{g(x)}{x}=a
$$
and $f^{(k)}(0)=g^{(k)}(0)$ for $1<k<n$, but $f^{(n)}(0)\not=g^{(n)}(0)$, then
$$
\lim_{x\to0}\frac{f(x)-g(x)}{g^{-1}(x)-f^{-1}(x)}=a^{n+1}
$$

Simpler approach: Convinced that there must be a simpler approach, I have revisited this answer.

For some $n>1$, assume


*

*$f,g\in C^n$

*$f^{(k)}(0)=g^{(k)}(0)$ for $k< n$ and $f^{(n)}(0)\not=g^{(n)}(0)$

*$f(0)=0$ and $f'(0)=a\not=0$
These assumptions imply that $f(x)=ax+O(x^2)$ and $g(x)=ax+O(x^2)$.
Furthermore, $f^{-1}(x)=x/a+O(x^2)$ and $g^{-1}(x)=x/a+O(x^2)$.
Assumption 2 implies that
$$
\lim_{x\to0}\frac{f(x)-g(x)}{x^n}=\frac{f^{(n)}(0)-g^{(n)}(0)}{n!}\not=0\tag{6}
$$
Substituting $x\mapsto f^{-1}(x)$ and using $\lim\limits_{x\to0}\frac{g(g^{-1}(x))-g(f^{-1}(x))}{g^{-1}(x)-f^{-1}(x)}=g'(0)=a$ yields
$$
\begin{align}
\lim_{x\to0}\frac{f(x)-g(x)}{x^n}
&=\lim_{x\to0}\frac{f(f^{-1}(x))-g(f^{-1}(x))}{f^{-1}(x)^n}\\
&=\lim_{x\to0}\frac{g(g^{-1}(x))-g(f^{-1}(x))}{f^{-1}(x)^n}\\
&=\lim_{x\to0}\frac{a(g^{-1}(x)-f^{-1}(x))}{(x/a)^n}\\
&=a^{n+1}\lim_{x\to0}\frac{g^{-1}(x)-f^{-1}(x)}{x^n}\tag{7}
\end{align}
$$
Dividing both sides of $(7)$ by $\lim\limits_{x\to0}\frac{g^{-1}(x)-f^{-1}(x)}{x^n}$ yields
$$
\lim_{x\to0}\frac{f(x)-g(x)}{g^{-1}(x)-f^{-1}(x)}=a^{n+1}\tag{8}
$$
A: By Taylor's series we obtain that

*

*$\sin (\tan x)= \sin\left(x + \frac13 x^3 + \frac2{15}x^5+ \frac{17}{315}x^7+O(x^9) \right)=x + \frac16 x^3 -\frac1{40}x^5 - \frac{55}{1008}x^7+O(x^9)$
and similarly

*

*$\tan (\sin x)= x + \frac16 x^3 -\frac1{40}x^5 - \frac{107}{5040}x^7+O(x^9)$

*$\arcsin (\arctan x)= x - \frac16 x^3 +\frac{13}{120}x^5 - \frac{341}{5040}x^7+O(x^9)$

*$\arctan (\arcsin x)= x - \frac16 x^3 +\frac{13}{120}x^5 - \frac{173}{5040}x^7+O(x^9)$
which leads to the result.
