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 $(x^2(\arcsin{\frac{1}{x}} -\arcsin{\frac{x}{1+x}} )= 1$
 A: 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.
A: 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.
A: 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.
A: Its much easier to use the formula $$\arcsin a - \arcsin b = \arcsin(a\sqrt{1 - b^{2}} - b\sqrt{1 - a^{2}})$$ directly. Clearly if $a = 1/x$ and $b = 1/(x + 1)$ we see that
\begin{align}
A &= a\sqrt{1 - b^{2}} - b\sqrt{1 - a^{2}}\notag\\
&= \frac{\sqrt{x^{2} + 2x}}{x(x + 1)} - \frac{\sqrt{x^{2} - 1}}{x(x + 1)}\notag\\
&= \frac{(x^{2} + 2x) - (x^{2} - 1)}{x(x + 1)\{\sqrt{x^{2} + 2x} + \sqrt{x^{2} - 1}\}}\notag\\
&= \frac{2x + 1}{x(x + 1)\{\sqrt{x^{2} + 2x} + \sqrt{x^{2} - 1}\}}\notag\\
\end{align}
Clearly we can see that $A \to 0$ as $x \to \infty$ and hence $$\lim_{x \to \infty}x^{2}\arcsin A = \lim_{x \to \infty}Ax^{2}\cdot\frac{\arcsin A}{A} = \lim_{x \to \infty}Ax^{2}$$ and we have
\begin{align}
Ax^{2} &= \frac{x^{2}(2x + 1)}{x(x + 1)\{\sqrt{x^{2} + 2x} + \sqrt{x^{2} - 1}\}}\notag\\
&=\dfrac{2 + \dfrac{1}{x}}{\left(1 + \dfrac{1}{x}\right)\left(\sqrt{1 + \dfrac{2}{x}} + \sqrt{1 - \dfrac{1}{x^{2}}}\right)}\notag\\
&\to \frac{2}{1\cdot(1 + 1)} = 1\text{ as }x \to \infty\notag
\end{align}
A: Noting that $\frac{d}{dx}\arcsin\left(\frac{1}{x}\right)=\frac{-1}{x\sqrt{x^2-1}}$, we can write:
$$x^2\left(\arcsin\left(\frac{1}{x}\right)-\arcsin\left(\frac{1}{x+1}\right)\right)=x^2\int_x^{x+1}\frac{dt}{t\sqrt{t^2-1}}$$
The integrand is nestled between $\frac{1}{(x+1)\sqrt{x^2+2x}}$ and $\frac{1}{x\sqrt{x^2-1}}$, both of which are readily asymptotic to $\frac{1}{x^2}$. That the limit is $1$ follows directly from this by considering the squeeze theorem
