Compute the following limit: $\lim_{n \to \infty}{\frac{\arcsin{\frac{1}{n}}-\arctan{\frac{1}{n}}}{\sin{\frac{1}{n}}-\tan{\frac{1}{n}}}}$ I'm trying to solve the following limit:
$\displaystyle \lim_{n \to \infty}{\frac{\arcsin{\frac{1}{n}}-\arctan{\frac{1}{n}}}{\sin{\frac{1}{n}}-\tan{\frac{1}{n}}}}$, but I do not have any idea about how to start. 
Any help will be very useful.
 A: Does the limit$$\lim_{h\to0}\frac{\arcsin h-\arctan h}{\sin h-\tan h}$$exist? If it does, it will be equal to your limit. But\begin{align}\lim_{h\to0}\frac{\arcsin h-\arctan h}{\sin h-\tan h}&=\lim_{h\to0}\frac{\left(h+\frac{h^3}6+\frac{3h^5}{40}+\cdots\right)-\left(h-\frac{h^3}3+\frac{h^5}5+\cdots\right)}{\left(h-\frac{h^3}6+\frac{h^5}{120}+\cdots\right)-\left(h+\frac{h^3}3+\frac{2h^5}{15}+\cdots\right)}\\&=\lim_{h\to0}\frac{\frac{h^3}2-\frac{h^5}8+\cdots}{-\frac{h^3}2-\frac{h^5}8+\cdots}\\&=-1.\end{align}
A: You may want to compute
$$
\lim_{x\to0}\frac{\arcsin x-\arctan x}{\sin x-\tan x} \tag{*}
$$
which your limit will be equal to, provided (*) exists.
The limit of the sequence might exist also if the limit of the function doesn't. But if the limit of the function exists, also does the limit of the sequence and they're equal.
Evaluating the order of infinitesimal of the denominator helps:
$$
\sin x-\tan x=\sin x\frac{\cos x-1}{\cos x}
$$
With Taylor expansion:
$$
\sin x\frac{\cos x-1}{\cos x}=
\frac{1}{\cos x}(x+o(x))\left(-\frac{x^2}{2}+o(x^2)\right)=
\frac{1}{\cos x}\left(-\frac{x^3}{2}+o(x^3)\right)
$$
The cosine can be left as is, because it doesn't contribute.
Thus we know that we have to expand the numerator up to degree $3$. The Taylor expansion of the arctangent is
$$
\arctan x=x-\frac{x^3}{3}+o(x^3)
$$
For the arcsine, we can consider
$$
\arcsin'x=(1-x^2)^{-1/2}=1+\frac{x^2}{2}+o(x^2)
$$
and so
$$
\arcsin x=x+\frac{x^3}{6}+o(x^3)
$$
Then
$$
\lim_{x\to0}\frac{\arcsin x-\arctan x}{\sin x-\tan x}=
\lim_{x\to0}\cos x\frac{(x+x^3/6)-(x-x^3/3)+o(x^3)}{-x^3/2+o(x^3)}=-1
$$
because
$$
\frac{1}{6}+\frac{1}{3}=\frac{1}{2}
$$
A: If $\arctan h=u,\sin u=h,\cos u=\sqrt{1-h^2},\tan u=\dfrac h{\sqrt{1-h^2}},u=\arctan\dfrac h{\sqrt{1-h^2}}$
$$\dfrac{\arcsin h-\arctan h}{\sin h-\tan h}=-\cos h\cdot\dfrac{\arctan\dfrac h{\sqrt{1-h^2}}-\arctan h}{\sin h(1-\cos h)}$$
Now $\arctan\dfrac h{\sqrt{1-h^2}}-\arctan h=\arctan\dfrac{\dfrac h{\sqrt{1-h^2}}-h}{1+\dfrac{h^2}{\sqrt{1-h^2}}}$
$=\arctan\dfrac{h(1-\sqrt{1-h^2})}{\sqrt{1-h^2}+h^2}=\arctan\dfrac{h^3}{(\sqrt{1-h^2}+h^2)(1+\sqrt{1-h^2})}$
So, $\displaystyle\lim_{h\to0}\dfrac{\arctan\dfrac{h^3}{(\sqrt{1-h^2}+h^2)(1+\sqrt{1-h^2})}}{h^3}=\cdots=\dfrac1{1+\sqrt1}=?$
and $\displaystyle\lim_{h\to0}\dfrac{\sin h(1-\cos h)}{h^3}=\lim_{h\to0}\dfrac1{(1+\cos h)}\left(\lim_{h\to0}\dfrac{\sin h}h\right)^3=\dfrac1{1+\cos0}=?$
A: *

*A good point to start is always to put in some values (n=10, n=100, n=1000, ...) to get a first guess.

*In this particular case, try to look at the limit $1/n = x \rightarrow 0$ and apply L'Hôpital's rule (multiple times)
