<s>[INST] Proof of limit of $x^2\left(\arcsin\left(\frac{1}{x}\right)-\arcsin\left(\frac{1}{x+1}\right)\right)$ correct? I want to show that $$\lim_{x\to\infty} x^2\left(\arcsin\left(\frac{1}{x}\right)-\arcsin\left(\frac{1}{x+1}\right)\right)= 1  $$
But I'm not sure if it's correct:
Because $\arcsin:[-1;1]\to \mathbb{R}$ is continuous and differentiable on $(-1,1)$, the mean value theorem says that there is a $c$ in $\left(\frac{1}{x+1} , \frac{1}{x}\right)$ for $f(t)=\arcsin(t)$ such that
$$x^2\left(\arcsin\left(\frac{1}{x}\right)-\arcsin\left(\frac{1}{x+1}\right)\right)= f'(c_x) {\frac{x}{1+x}}  $$
then
$f'(c_x)=\frac{1}{(\sqrt{1-c_x})}(\frac{x}{x+1}))\leqslant \frac{1}{(\sqrt{1-c_x})}\times\frac{x}{x} = \frac{1}{(\sqrt{1-c_x})} $
with Sandwichlemma for $x\to\infty, c_x\to0$
so the limit of $(x^2(\arcsin{\frac{1}{x}} -\arcsin{\frac{x}{1+x}} )= 1$
 [/INST] You need to bound the function on both sides, one side is not enough. I'd proceed in a slightly different way, for applying the mean value theorem.
Consider the function
$$
f(x)=\arcsin\frac{1}{x}
$$
defined for $x>1$ (it is defined on a larger set, but $(1,\infty)$ is sufficient for our purposes).
By the mean value theorem, there exists $c_x\in(x,x+1)$ such that
$$
\frac{f(x)-f(x+1)}{x-(x+1)}=f'(c_x)
$$
so
$$
x^2(f(x)-f(x+1))=-x^2f'(c_x)
$$
Now
$$
f'(x)=\frac{1}{\sqrt{1-(1/x)^2}}\left(-\frac{1}{x^2}\right)=
-\frac{1}{x\sqrt{x^2-1}}
$$
so we end up with
$$
x^2(f(x)-f(x+1))=\frac{x^2}{c_x\sqrt{c_x^2-1}}
$$
Let's consider
$$
g(t)=t\sqrt{t^2-1}
$$
with
$$
g'(t)=\sqrt{t^2-1}+\frac{t^2}{\sqrt{t^2-1}}>0
$$
for $t>1$. Thus we can say, from $x<c_x<x+1$, that
$$
g(x)<g(c_x)<g(x+1)
$$
so
$$
\frac{x^2}{g(x+1)}\le x^2(f(x)-f(x+1))\le\frac{x^2}{g(x)}
$$
which is the same as
$$
\frac{x^2}{(x+1)\sqrt{(x+1)^2-1}}\le 
x^2(f(x)-f(x+1))\le
\frac{x^2}{x\sqrt{x^2-1}}
$$
The squeeze theorem allows you to conclude.
</s>