Need help with Limit solving

Question

How do I evaluate this limit?

$$\lim_{x \to 0}\frac{\cos\sin\tan x - \cos\tan\sin x}{\cosh\arcsin x + \cosh\sin x - 2\cosh x}$$

I don't really have ideas on how to start. I think L'Hopital rule does not help in this case.

Answer 1

As suggested in the comments, the main difficulty with this sort of problems is where to stop in taking the Taylor expansion of the terms.

Asymptotics and de L'Hopital rule do not help, the first just doesn't work, the second takes too long, especially for the numerator.

In our try to solve this limit we will use the following Taylor expansions, centered in $\ x_0=0$, up to the sixth power (in the first two for example, there is not the term to the sixth power, so we stop at the fifth):

$$\ \sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}+o(x^6)$$ $$\ \tan(x)=x+\frac{x^3}{3}+\frac{2}{15}x^5+o(x^6)$$ $$\ \cos(x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+o(x^6)$$ $$\ \arcsin(x)=x+\frac{x^3}{6}+\frac{3}{40}x^5+o(x^6)$$ $$\ \cosh(x)=1+\frac{x^2}{2!}+\frac{x^4}{4!}+\frac{x^6}{6!}+o(x^6)$$

Other rules to remember, $\ \forall m,n>0$: $$\ o(x^m)\pm o(x^n)=o(x^{min[n,m]})$$ $$\ x^m\cdot o(x^n)=o(x^{m+n})$$

Now we can begin: $$\ \color{red}{\sin(\tan(x))}=\sin(x+\frac{x^3}{3}+\frac{2}{15}x^5+o(x^6))=$$ $$\ =x+\frac{x^3}{3}+\frac{2}{15}x^5+o(x^6)-\frac{1}{6}\bigg[x+\frac{x^3}{3}+\frac{2}{15}x^5+o(x^6)\bigg]^3+\frac{1}{120}x^5+o(x^6)$$

where in the last part raised to the fifth power I just took the term that I need, because all the others have an exponent which is greater than 6.

Now let us examine the term raised to the third power: the $\ o(x^6)$ both raised to the third power or multiplied by any of the other terms will give something which is even smaller than itself, so I am not writing it anymore in the squared brackets; and the same happens for the $\ \frac{2}{15}x^5$ term:

$$\ \bigg[x+\frac{x^3}{3}\bigg]^3=x^3+x^5+\underbrace{\frac{x^7}{2}+\frac{x^9}{27}}_{o(x^6)}$$ So we have: $$\ \color{red}{\sin(\tan(x))}=x+\frac{x^3}{3}+\frac{2}{15}x^5-\frac{1}{6}\bigg(x^3+x^5+o(x^6)\bigg)+\frac{x^5}{120}+o(x^6)=$$ $$\ =x+\frac{x^3}{3}+\frac{2}{15}x^5-\frac{x^3}{6}-\frac{x^5}{6}+\frac{x^5}{120}+o(x^6)=$$ $$\ =x+\frac{x^3}{6}+x^5\bigg(\frac{16-20+1}{120}\bigg)+o(x^6)=$$ $$\ =\color{red}{x+\frac{x^3}{6}-\frac{x^5}{40}+o(x^6)}$$

In a similar way, we can find that $$\ \color{blue}{\tan(\sin(x))}=\tan(x-\frac{x^3}{6}+\frac{x^5}{120}+o(x^6))=$$ $$\ =x-\frac{x^3}{6}+\frac{x^5}{120}+o(x^6)+\frac{1}{3}\bigg[x-\frac{x^3}{6}+\frac{x^5}{120}+o(x^6)\bigg]^3+\frac{2}{15}x^5+o(x^6)=$$ $$\ =\color{blue}{x+\frac{x^3}{6}-\frac{x^5}{40}+o(x^6)}$$ So we showed that up to the sixth degree they have the same expansion, and this means that: $$\ \cos(\sin(\tan x))-\cos(\tan(\sin x))=o(x^6)$$ Now we must deal with the denominator, hoping to get a term greater that $\ x^6$ , otherwise I would still have an indetermined form and I should start again from the beginning, adding another term in the expansions above. $$\ \color{magenta}{\cosh(\arcsin x)}=\cosh(x+\frac{x^3}{6}+\frac{3}{40}x^5+o(x^6))=$$ $$\ =1+\frac{1}{2}\bigg[x+\frac{x^3}{6}+\frac{3}{40}x^5+o(x^6)\bigg]^2+\frac{1}{24}\bigg[x+\frac{x^3}{6}+\frac{3}{40}x^5+o(x^6)\bigg]^4+$$ $$\ +\frac{1}{720}\bigg[x+\frac{x^3}{6}+\frac{3}{40}x^5+o(x^6)\bigg]^6+o(x^6)=$$ $$\ =1+\frac{1}{2}\bigg[x^2+\frac{x^4}{3}+\frac{x^6}{36}+\frac{3}{20}x^6+o(x^6)\bigg]+\frac{1}{24}\bigg[x^4+\frac{2}{3}x^6+o(x^6)\bigg]+\frac{1}{720}x^6+o(x^6)=$$ $$\ =1+\frac{x^2}{2}+\frac{5}{24}x^4+x^6\bigg(\frac{1}{72}+\frac{3}{40}+\frac{1}{36}+\frac{1}{720}\bigg)+o(x^6)=$$ $$\ =\color{magenta}{1+\frac{x^2}{2}+\frac{5}{24}x^4+\frac{17}{144}x^6+o(x^6)}$$ $$\ $$ $$\ \color{Green}{\cosh(\sin x)}=\cosh(x-\frac{x^3}{6}+\frac{x^5}{120}+o(x^6))=$$ $$\ =1+\frac{1}{2}\bigg[x-\frac{x^3}{6}+\frac{x^5}{120}+o(x^6)\bigg]^2+\frac{1}{24}\bigg[x-\frac{x^3}{6}+\frac{x^5}{120}+o(x^6)\bigg]^4+\frac{1}{720}x^6+o(x^6)=$$ $$\ =1+\frac{x^2}{2}-\frac{x^4}{8}+x^6\bigg(\frac{1}{45}-\frac{1}{36}+\frac{1}{720}\bigg)+o(x^6)=$$ $$\ =\color{Green}{1+\frac{x^2}{2}-\frac{x^4}{8}-\frac{x^6}{240}+o(x^6)}$$

Therefore, the denominator is: $$\ \color{Goldenrod}{\cosh(\arcsin x)+\cosh(\sin x)-2\cosh x}=$$ $$\ =1+\frac{x^2}{2}+\frac{5}{24}x^4+\frac{17}{144}x^6+1+\frac{x^2}{2}-\frac{x^4}{8}-\frac{x^6}{240}+$$ $$\ -2\bigg(1+\frac{x^2}{2}+\frac{x^4}{24}+\frac{x^6}{720}+o(x^6)\bigg)+o(x^6)=$$ $$\ =\bigg(1+1-2\bigg)+x^2\bigg(\frac{1}{2}+\frac{1}{2}-1\bigg)+x^4\bigg(\frac{5-3-2}{24}\bigg)+x^6\bigg(\frac{85-3-2}{720}\bigg)+o(x^6)=$$ $$\ =\color{Goldenrod}{\frac{x^6}{9}+o(x^6)}$$ Finally: $$\ \lim_{x\to0}\frac{\cos(\color{red}{\sin(\tan x)})-\cos(\color{blue}{\tan(\sin x)})}{\color{Goldenrod}{\cosh(\arcsin x)+\cosh(\sin x)-2\cosh x}}=$$ $$\ =\lim_{x\to0}\frac{o(x^6)}{\frac{x^6}{9}+o(x^6)}=\lim_{x\to0}\frac{o(x^6)}{x^6(\frac{1}{9}+o(1))}=0$$