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How would I find this limit?

$\lim_{n \to \infty} \frac{\sqrt{n}}{2} \bigl(\arccos(\frac{n-2}{22+n})) $

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up vote 3 down vote accepted

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: $$ \begin{eqnarray} \lim_{n \to \infty} \frac{ \arccos\left( \frac{1-2/n}{1+22/n} \right)}{2 n^{-1/2}} &=& \lim_{n \to \infty} \left(-n^{3/2} \right) \left(-\frac{2 \sqrt{3}}{\sqrt{n+10} (n+22)} \right) \\ &=& 2 \sqrt{3} \lim_{n \to \infty} \frac{1}{\sqrt{1+\frac{10}{n}} \left( 1 + \frac{22}{n} \right)} = 2 \sqrt{3} \end{eqnarray}$$

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Thank you for useful decision) –  Sh.N. May 17 '12 at 17:59
    
What happened to -n^\frac{3}{2} ? Was it factored out? –  Larry Battle May 17 '12 at 18:28
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@LarryBattle $n^{3/2}$ was rewritten as a product $\sqrt{n} \cdot n$, and then combined using $\frac{\sqrt{n}}{\sqrt{n+10}} \cdot \frac{n}{n+22} = \frac{1}{\sqrt{1+10/n}} \cdot \frac{1}{1+22/n}$ –  Sasha May 17 '12 at 18:30
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