How to show that for $-1\leq x_n \leq 1$ and $\lim_{n \to \infty} x_n = a$, we have $\lim_{n \to \infty} \arcsin x_n = \arcsin a$? 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.
 A: 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 $\epsilon \gt 0$.
A: 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, $|f^{-1}(s) - f^{-1}(t)| \le 2 (b-a)/n < \epsilon$.
