Please help me evaluate limit of $x$ tending towards the dottie number

Question

$$\lim _{\frac{x}{\cos x}→1}\left(\frac{ \underbrace{\arccos \circ \arccos \circ \ldots \circ \arccos (x)}_{n~\text{iterations}} -\underbrace{\cos \circ \cos \circ \ldots \circ \cos (x)}_{n~\text{iterations}}}{\left(x-\cos x\right)}\right)$$

as ${\frac{x}{\cos x}→1}$, I suppose that $x$ tends towards the Dottie number, the unique real root of this equation : $\cos(x)=x, \,x≈0.73908513321516064165531208767\ldots$.
More about it here-- How to compute Dottie number accurately?)

It is a $0/0$ form, yet the application of L'Hopital's rule becomes difficult, as iterations of function depend upon $n$, which isn't known, or has any condition whatsoever. I have taken the derivative of the denominator, which turns out to be $1+\sin(x)$, but I have no idea on how to proceed for the numerator, especially the $\arccos$ part . Please help me to reach the solution. Thanks in advance.
Note- here $n$ represents n iterations of cos and arccosine, eg. at $n=2$ numerator would be $\arccos(\arccos(x))-\cos(\cos(x))$.

Answer 1

The hardest part of this problem was developing a good notation. I also used LHR but there might be a way to avoid it.

Let $f^{(n)}$ denote $n$-fold composition. Let $c$ denote the Dottie number, i.e. the unique real solution of $x=\cos(x)$. The condition $x/\cos(x)\to 1$ is equivalent to the condition $x\to c$. Note that $1-c^2=\sin^2(c)$, and in particular that $(1-c^2)^{-1/2}=\csc(c)$.

Claim: $$ \lim_{x\to c}\frac{\arccos^{(n)}(x)-\cos^{(n)}(x)}{x-\cos(x)}=(-1)^n \frac{\csc^n(c)-\sin^n(c)}{1+\sin(c)} $$Proof: The quickest way I see to evaluate this limit is LHR: The limit is indeterminate as $x\to c$ for every $n$ and everything is differentiable, so we're justified in using it. $$ \lim_{x\to c}\frac{\arccos^{(n)}(x)-\cos^{(n)}(x)}{x-\cos(x)}= \lim_{x\to c}\frac{\arccos\left(\arccos^{(n-1)}(x)\right)-\cos\left(\cos^{(n-1)}(x)\right)}{x-\cos(x)} $$ $$ \stackrel{LHR}{=}\lim_{x\to c}\frac{-(1-\cos^{(n-1)}(x))^{-1/2}\cdot \frac{d}{dx}\arccos^{(n-1)}(x)-(-\sin(\cos^{(n-1)}(x))\cdot \frac{d}{dx}\cos^{(n-1)}(x))}{1+\sin(x)}= $$ $$ =\lim_{x\to c}\frac{\prod\limits_{k=1}^{n}-(1-\cos^{(n-k)}(x))^{-1/2}-\prod\limits_{k=1}^{n}(-\sin(\cos^{(n-k)}(x)))}{1+\sin(x)}= $$ $$ =(-1)^n\lim_{x\to c}\frac{\prod\limits_{k=1}^{n}(1-\cos^{(n-k)}(x))^{-1/2}-\prod\limits_{k=1}^{n}(\sin(\cos^{(n-k)}(x)))}{1+\sin(x)} $$ Now we can take the limit; note that we can replace all instances of $\cos(c)$ with $c$. $$ =(-1)^n \frac{(\csc(c))^n -(\sin(c))^n}{1+\sin(c)} $$I checked and this matches the numerical values for the first $10$ cases.

Answer 2

For any function $f$ and any $n\in\mathbb{N}$, define $f_n$ as

$$f_n(x)= \begin{cases} x&\text{if }n=0\\ f(f_{n-1}(x))&\text{if }n\gt0 \end{cases}$$

Note that, if $f$ is differentiable, the Chain Rule tells us $f_n'(x)=f'(f_{n-1}(x))f_{n-1}'(x)$ for $n\gt0$.

Now let $F(x)=\arccos x$ and $G(x)=\cos x$, for which $F'(x)=-1/\sqrt{1-x^2}$ and $G'(x)=-\sin x$. The OP's problem is to compute

$$\lim_{x\to x_0}{F_n(x)-G_n(x)\over x-\cos x}$$

where $x_0$ is the "Dottie" number, which satisfies $x_0=\cos x_0$. We see that

$$F(x_0)=\arccos x_0=\arccos(\cos x_0)=x_0$$

so that, by induction,

$$F_n'(x_0)=F'(F_{n-1}(x_0))F_{n-1}'(x_0)=\left(-1\over\sqrt{1-x_0^2}\right)^n$$

and

$$G(x_0)=\cos x_0=x_0$$

so that, again by induction,

$$G_n'(x_0)=G'(G_{n-1}(x_0))G_{n-1}'(x_0)=(-\sin x_0)^n=\left(-\sqrt{1-x_0^2}\right)^n$$

L'Hopital thus tells us

$$\lim_{x\to x_0}{F_n(x)-G_n(x)\over x-\cos x}=\lim_{x\to x_0}{F_n'(x_0)-G_n'(x_0)\over1+\sin x_0}=(-1)^n{1-(1-x_0^2)^n\over(1-x_0^2)^{n/2}+(1-x_0^2)^{(n+1)/2}}$$