How to find limit of function: $\lim\limits_{n \to \infty} \frac{\sqrt{n}}{2} \arccos(\frac{n-2}{22+n})$ How would I find this limit?
$\lim_{n \to \infty} \frac{\sqrt{n}}{2} \bigl(\arccos(\frac{n-2}{22+n}))$

As the argument of $\arccos$ approaches 1 for large $n$, the inverse cosine function approaches zero. Rewriting the expression as quotient of two expression approaching zero for 
large $n$ 
$$
   \frac{ \arccos\left( \frac{1-2/n}{1+22/n} \right)}{2 n^{-1/2}} 
$$
we can apply the L'Hospital's rule:
$$ \begin{eqnarray}
 \lim_{n \to \infty} \frac{ \arccos\left( \frac{1-2/n}{1+22/n} \right)}{2 n^{-1/2}}  &=&
   \lim_{n \to \infty} \left(-n^{3/2} \right) \left(-\frac{2 \sqrt{3}}{\sqrt{n+10} (n+22)} \right) \\ &=& 2 \sqrt{3} \lim_{n \to \infty} \frac{1}{\sqrt{1+\frac{10}{n}} \left( 1 + \frac{22}{n} \right)} = 2 \sqrt{3}
\end{eqnarray}$$
