<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] If one wishes to use the MVT, then for $f(x)=\arcsin(x)$, there exists $\xi\in (y,z)$ such that
$$\arcsin(z)-\arcsin(y)=\frac{1}{\sqrt{1-\xi^2}}(z-y)$$
Letting $z=\frac1x$ and $y=\frac1{x+1}$, we have
$$x^2\left(\arcsin\left(\frac{1}{x}\right)-\arcsin\left(\frac{1}{x+1}\right)\right)=\frac{1}{\sqrt{1-\xi^2}}\frac{x}{x+1}$$
Inasmuch as $\frac{1}{x+1}<\xi<\frac1x$, we have $\lim_{x\to \infty}\xi =0$ and therefore,
$$\lim_{x\to \infty}\left(x^2\arcsin\left(\frac{1}{x}\right)-x^2\arcsin\left(\frac{1}{x+1}\right)\right)=1$$
If one would like to use the squeeze theorem in a more direct manner, then we have 
$$\frac{x+1}{\sqrt{x(x+2)}}<\frac{1}{\sqrt{1-\xi^2}}<\frac{x}{\sqrt{x^2-1}}$$
which shows that 
$$\frac{x}{\sqrt{x(x+2)}}<\left(x^2\arcsin\left(\frac{1}{x}\right)-x^2\arcsin\left(\frac{1}{x+1}\right)\right)<\frac{x^2}{(x+1)\sqrt{x^2-1}}$$
whereupon application of the squeeze theorem yields the expected result.
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