Find the limit $\displaystyle{\lim_{x\to 0}\frac{\sin(x)\sin^{-1}(x)-\sinh(x)\sinh^{-1}(x)}{x^2(\cos(x)-\cosh(x)+\sec(x)-\text{sech}(x))}}$

Question

Here is an example from the book "Asymptotic Analysis and Perturbation Theory" by William Paulsen.

Example 1.10 p.18.

Find $$\lim_{x\to 0}\frac{\sin(x)\sin^{-1}(x)-\sinh(x)\sinh^{-1}(x)}{x^2(\cos(x)-\cosh(x)+\sec(x)-\text{sech}(x))}$$

As this is an asymptotic book, the solution in the book uses the Maclaurin expansion of all the trig functions and simplifies the denominator to $x^8/6+\mathcal{O}(x^{10})$. Then keeping just the right number of terms, the numerator is computed as $x^8/15+\mathcal{O}(x^{10})$. Finally, $$\frac{\sin(x)\sin^{-1}(x)-\sinh(x)\sinh^{-1}(x)}{x^2(\cos(x)-\cosh(x)+\sec(x)-\text{sech}(x))}\sim\frac{x^8/15+\mathcal{O}(x^{10})}{x^8/6+\mathcal{O}(x^{10})}=\frac{2}{5}+\mathcal{O}(x^2)$$ I wonder if there are any ways to find the higher terms in the asymptotics or just any other ways to evaluate the limit. Thanks in advance.

Answer 1

This has not been done with pen and paper. $$A=\sin ^{-1}(x) \sin (x)-\sinh ^{-1}(x) \sinh (x)$$ $$A==\frac{x^8}{15}+\frac{601 x^{12}}{17010}+\frac{16423637 x^{16}}{729729000}+\frac{9557991941 x^{20}}{601184430000}+O\left(x^{24}\right)$$ $$B=\cos (x)-\cosh (x)+\sec (x)-\text{sech}(x)$$ $$B=\frac{x^6}{6}+\frac{421 x^{10}}{15120}+\frac{4057 x^{14}}{887040}+\frac{53584663 x^{18}}{71327692800}+\frac{21014810422163 x^{22}}{170303140572364800}+O\left(x^{24}\right)$$

Now, long division to obtain $$\frac{A}{x^2\,B}=\frac{2}{5}+\frac{8231 x^4}{56700}+\frac{4078718041 x^8}{40864824000}+\frac{25535089927517 x^{12}}{350129812032000}+O\left(x^{16}\right)$$

Answer 2

The denominator can be simplified as $$x^2(\cos x - \cosh x) \cdot\frac{\cos x\cosh x-1}{\cos x\cosh x} $$ and the denominator in above expression tends to $1$ and can be safely replaced by $1$ so that effectively we can write denominator of original expression as $$x^2(\cos x-\cosh x) (\cos x\cosh x-1)\tag{1}$$ Next we can observe that both $\frac {1-\cos x} {x^2},\frac{\cosh x-1}{x^2}$ tend to $1/2$ and hence their sum tends to $1$. In other words the expression $$\frac{\cosh x-\cos x} {x^2}\to 1$$ And then the expression $(1)$ can be replaced by $$x^4(1-\cos x\cosh x) \tag{2}$$ Let us now observe that $$(1-\cos x) (1-\cosh x) =2-\cos x-\cosh x-(1-\cos x\cosh x) $$ and the left side when divided by $x^4$ tends to $-1/4$ and hence we have $$\lim_{x\to 0}\frac{1-\cos x\cosh x} {x^4}=\frac{1}{4}+\lim_{x\to 0}\frac{2-\cos x-\cosh x} {x^4}$$ One can now apply Taylor or l'Hospital's Rule to conclude that $$\lim_{x\to 0}\frac{1-\cos x\cosh x} {x^4}=\frac{1}{6}$$ It follows from above equation and $(2)$ that the denominator of original expression under limit can be replaced by $x^8/6$.

The handling of numerator is a bit more involved and I doubt if one can really avoid Taylor expansions here. Let us observe that the numerator can be written as $f(x) +f(ix) $ where $f(x) =\sin x\sin^{-1}x$ is an even function with the series expansion $$f(x) =ax^2+bx^4+cx^6+dx^8+\dots$$ and then $$f(x) +f(ix) = 2bx^4+2dx^8+\dots$$ so that we only need to evaluate the coefficients of $x^{4n}$ in series for $f(x) $. Given the series expansions $$\sin x=x-\frac{x^3}{6}+\frac{x^5}{120}-\frac{x^7}{5040}+\dots$$ and $$\sin^{-1}x=x+\frac{x^3}{6}+\frac{3x^5}{40}+\frac{5x^7}{112}+\cdots$$ we get $$b=0,d=\frac{5}{112}-\frac{1}{80}+\frac{1}{720}-\frac{1}{5040}=\frac{1}{30}$$ and hence the numerator can be written as $$\frac{x^8}{15}+o(x^8)$$ The desired limit is thus $6/15=2/5$.