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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.

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Can't say much about the limit, but your quote is attributed to G.H. Hardy in "A Mathematician's Apology". –  Austin Mohr Nov 8 '11 at 22:52
@Mohr: Thanks! ${}$ –  Jonas Teuwen Nov 8 '11 at 22:52
This might of interest: both the numerator and the denominator are $(-\frac{1}{30}) \ x^7 + O(x^9)$ by Taylor expansion. –  Srivatsan Nov 8 '11 at 22:58
This is item #2 in Arnold's Trivium (French version, translated from Russian Math. Surveys 46:1 (1991)). –  Did Nov 8 '11 at 23:03
Yes, I was wondering whether Jonas was having dinner with Arnold! –  Gerry Myerson Nov 9 '11 at 3:46

2 Answers 2

up vote 12 down vote accepted

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

  1. $f,g\in C^n$

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

  3. $f(0)=0$ 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 $$ \lim_{x\to0}\frac{f(x)-g(x)}{g^{-1}(x)-f^{-1}(x)}=a^{n+1}\tag{8} $$

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Well done! This is a L'Hopital solution, but then nicer than the one I had. One could also fill in odd-degree polynomials (since that will be the series expansion) and get the result. I wonder if there is a nicer trick. –  Jonas Teuwen Nov 9 '11 at 12:27
+1 A very nice approach! Quite concise, elegant and to the point. –  Domagoj Pandža Apr 4 '12 at 17:56

See the discussion at MathOverflow.

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