Question: How to show that for $-1\leq x\_n \leq 1$ and $\lim\_{n \to \infty} x\_n = a$, we have $\lim\_{n \to \infty} \arcsin x\_n = \arcsin a$?

My solution: I tried to compute $\sin( \arcsin x\_n - \arcsin a )$. We have
$$
\sin( \arcsin x\_n - \arcsin a ) \\ 
= x\_n \cos(\arcsin a) - \cos(\arcsin x\_n)a \\
= x\_n \sqrt{1-a^2} - a \sqrt{1-x\_n^2}.
$$
Therefore $\lim\_{n \to \infty} \sin(\arcsin x\_n - \arcsin a) = 0$. It follows that $\lim\_{n \to \infty} (\arcsin x\_n - \arcsin a) = 0$. Hence $\lim\_{n \to \infty} \arcsin x\_n = \arcsin a$. Is this correct? Thank you very much. 

Edit: the proof should use only definition of a limit.

Answer: 
Things are fine up to 
$$x\_n\sqrt{1-a^2}-a\sqrt{1-x\_n^2}.$$
If $a=0$, we are essentially finished. It is also not very hard to deal with $a=\pm 1$. So let us suppose that $a\ne 0$, $a\ne \pm 1$. Then for large enough $n$, $x\_n$ has the same sign as $a$. 
Multiply top and bottom by $x\_n\sqrt{1-a^2}+a\sqrt{1-x\_n^2}$. We get
$$\frac{(x\_n-a)(1+ax\_n)}{x\_n\sqrt{1-a^2}+a\sqrt{1-x\_n^2}}.$$
Now we can, with a bit of effort, find a suitable $N$ for any $\epsilon \gt 0$.
**Theorem**: if $f$ is continuous and strictly increasing on $[a,b]$, then the inverse function $f^{-1}$ is continuous on $[f(a), f(b)]$.

*Proof*: Given $\epsilon > 0$, take a positive integer $n$ large enough that 
$(b-a)/n < \epsilon/2$. Let $x\_j = a + j (b-a)/n$ for $j = 0,1,\ldots, n$,
and $t\_j = f(x\_j)$. Since $a = x\_0 < x\_1 < \ldots < x\_n = b$, we have
$f(a) = t\_0 < t\_1 < \ldots < t\_n = f(b)$. Let $\delta = \min(t\_j - t\_{j-1})$. Then if $s,t \in [f(a),, f(b)]$ with $|s - t| < \delta$, $s$ and $t$ are either in the same interval $[t\_j, t\_{j+1}]$ or in neighbouring intervals $[t\_j,t\_{j+1}]$ and $[t\_{j+1},t\_{j+2}]$, and so $f^{-1}(s)$ and $f^{-1}(t)$ are either in the same interval $[x\_j, x\_{j+1}]$ or neighbouring intervals $[x\_j,x\_{j+1}]$ and $[x\_{j+1},x\_{j+2}]$. 
In particular, $|f^{-1}(s) - f^{-1}(t)| \le 2 (b-a)/n < \epsilon$.