Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How would I find this limit?

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

share|cite|improve this question

1 Answer 1

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

share|cite|improve this answer
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
@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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.