<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] Your proof is not complete: to use the Squeeze Theorem, you need both an upper and a lower bound converging to the same value (i.e., here $1$). But you only provided an upper bound.
Now, you have $c_x \in \left(\frac{1}{x+1}, \frac{1}{x}\right)$, so 
$$f^\prime(c_x) = \frac{1}{\sqrt{1-c_x^2}} \geq \frac{1}{\sqrt{1-\frac{1}{(x+1)^2}}}
$$
and therefore
$$f^\prime(c_x)\cdot \frac{x}{x+1} \geq \frac{1}{\sqrt{1-\frac{1}{(x+1)^2}}}\cdot \frac{x}{x+1} \xrightarrow[x\to\infty]{}1.
$$
(In your question, you appear to have missed the square in the derivative of $\arcsin$.) Also, note another mistake in what you wrote (for the upper bound): $c_x\xrightarrow[x\to\infty]{}0$, not to $\infty$, since $c_x \in \left(\frac{1}{x+1}, \frac{1}{x}\right)$.

For reference, using Taylor approximations: namely, we shall use the facts that, when $u\to 0$
$$\begin{align}
\arcsin u &= u + o(u^2) \\
\frac{1}{1+u} &= 1-u + o(u)
\end{align}$$
(Note that, as you'll see, we need the $o(u^2)$ in the first expansion, $o(u)$ is not enough.)${}^{(\dagger)}$

Taking $\frac{1}{x}\xrightarrow[x\to\infty]{} 0$ as "our $u$":
$$\begin{align}
\arcsin \frac{1}{x} &= \frac{1}{x} + o\!\left(\frac{1}{x^2}\right) \\
\arcsin \frac{1}{x+1} &= \arcsin \left(\frac{1}{x}\left(\frac{1}{1+\frac{1}{x}}\right)\right)
= \arcsin \left(\frac{1}{x}\left(1-\frac{1}{x} + o\!\left(\frac{1}{x}\right)  \right)\right)\\
&= \arcsin \left(\frac{1}{x}-\frac{1}{x^2} + o\!\left(\frac{1}{x^2}\right)\right)  \\&= \frac{1}{x}-\frac{1}{x^2} + o\!\left(\frac{1}{x^2}\right)
\end{align}$$
so that
$$\begin{align}
\arcsin \frac{1}{x}  - \arcsin \frac{1}{x+1} 
&=  \frac{1}{x} + o\!\left(\frac{1}{x^2}\right) - \left( \frac{1}{x}-\frac{1}{x^2} + o\!\left(\frac{1}{x^2}\right) \right) = \frac{1}{x^2} + o\!\left(\frac{1}{x^2}\right)
\end{align}$$
and to conclude
$$\begin{align}
x^2\left( \arcsin \frac{1}{x}  - \arcsin \frac{1}{x+1}  \right)
&=  x^2\left( \frac{1}{x^2} + o\!\left(\frac{1}{x^2}\right) \right) = 1 + o(1)
\xrightarrow[x\to\infty]{} 1.
\end{align}$$

$(\dagger)$ If we only used $\arcsin u = u+o(u)$, then we would get $\arcsin \frac{1}{x} = \frac{1}{x} + o\!\left(\frac{1}{x}\right)$ alright, but only $\arcsin \frac{1}{x+1} = \arcsin \left(\frac{1}{x}-\frac{1}{x^2} + o\!\left(\frac{1}{x^2}\right)\right) = \frac{1}{x}-\frac{1}{x^2} + o\!\left(\frac{1}{x^2}\right) + o\!\left(\frac{1}{x}\right) =\frac{1}{x} + o\!\left(\frac{1}{x}\right)$ as well. This is no longer sufficient to conclude, as now all we can say is
$$\begin{align}
x^2\left( \arcsin \frac{1}{x}  - \arcsin \frac{1}{x+1}  \right)
&=  x^2\left( o\!\left(\frac{1}{x}\right) \right) = o(x)
\end{align}$$
which could converge (or diverge) to basically anything.
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