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.
Best Answer
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 $6/15=2/5$.