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Find $$\lim\limits_{n \to \infty} \left( \frac{n^{1/3}}{2} \text{arccos} \left(\dfrac1{\sqrt{1+\frac{4}{(k(n)-1)^2}}\sqrt{1+\frac{8}{(k(n)-1)^2}}} \right) \right).$$

where $k(n) = \dfrac{1}{12} (108n+12 \sqrt{768+81n^2})^{1/3}-\dfrac{4}{ (108n+12 \sqrt{768+81n^2})^{1/3}} + 1$.

Help me, please.

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Have you tried a Taylor expansion of the arccos around the point 1? –  Eepzy May 21 '12 at 19:02
    
@Eepzy No, I haven't. –  Sh.N. May 21 '12 at 19:06
    
I tried to use L'Hopital's rule, but expressions are enormous. –  Sh.N. May 21 '12 at 19:55
    
I computated the function under 'lim' for large $n$. It is approximately $3\sqrt{2}$ (or $3.4$). –  Sh.N. May 21 '12 at 19:58
2  
@Sh.N. Could you tell us the source of this problem? –  user17762 May 24 '12 at 4:55

2 Answers 2

up vote 5 down vote accepted

The idea is to use l'Hospital rule in a judicious way. Before applying l'Hospital rule, let us do some algebraic manipulations and get it into a reasonably manageable form.

We need $$\lim\limits_{n \to \infty} \left( \frac{n^{1/3}}{2} \text{arcsec} \left(\sqrt{1+\frac{4}{k(n)^2}}\sqrt{1+\frac{8}{k(n)^2}} \right) \right).$$

where $k(n) = \dfrac{1}{12} (108n+12 \sqrt{768+81n^2})^{1/3}-\dfrac{4}{ (108n+12 \sqrt{768+81n^2})^{1/3}}$.

Note that my $k(n)$ is the OP's $k(n) - 1$. Also, note that we have converted the problem from $\arccos$ to $\text{arcsec}$ since $$\arccos \left( \dfrac1x \right) = \text{arcsec}(x)$$ Let $$f(n) = \text{arcsec} \left(\sqrt{1+\frac{4}{k(n)^2}}\sqrt{1+\frac{8}{k(n)^2}} \right)$$ and $$g(n) = \dfrac{2}{n^{1/3}}$$ We want $$\lim_{n \rightarrow \infty} \dfrac{f(n)}{g(n)}$$ We shall use L'Hospital rule. But before we jump into using L'Hospital rule, we will massage the expression so that it can be handled with relative easy.

$$\frac{df}{dn} = \dfrac{df}{dk} \dfrac{dk}{dn}$$

Recall that $$\dfrac{d \text{arcsec}(x)}{dx} = \dfrac1{x\sqrt{x^2-1}}.$$ Hence, $$\dfrac{df}{dk} = - \dfrac{k(6k^2+32)}{\sqrt{k^2+4}\sqrt{k^2+8}(k^2+4)(k^2+8)} \dfrac{\sqrt{k^4+12k^2+32}}{\sqrt{3k^2+8}}$$

$$\dfrac{dk}{dn} = \dfrac{3^{1/3} \left(3 \sqrt{3} n + \sqrt{256 + 27n^2} \right)\left(8 \times 3^{1/3} + 2^{1/3} \left( 9n + \sqrt{768+81n^2} \right)^{2/3} \right)}{2 \times 2^{2/3} \sqrt{256+27n^2} \left( 9n + \sqrt{768+81n^2} \right)^{4/3}}$$

$$\dfrac{dg}{dn} = - \dfrac{2}{3n^{4/3}}$$

Now note that as $n \rightarrow \infty$, $k(n) \rightarrow \infty$. As $k \rightarrow \infty$, $$\dfrac{df}{dk} \sim - 2\sqrt{3} \dfrac1{k^2}$$ As $n \rightarrow \infty$, $$\dfrac{dk}{dn} \sim \dfrac16 \frac1{n^{2/3}}$$

Hence, as $n \rightarrow \infty$, $$\dfrac{df}{dn} \sim - \dfrac1{\sqrt{3}} \frac1{k^2n^{2/3}}$$

When we write $h(n) \sim l(n)$, we mean that $\displaystyle \lim_{n \rightarrow \infty} \dfrac{h(n)}{l(n)} = 1$ i.e. $h(n) = l(n) + E(n)$, where $\dfrac{E(n)}{l(n)} \rightarrow 0$.

Hence, to summarize $$\lim\limits_{n \to \infty} \left( \frac{n^{1/3}}{2} \text{arcsec} \left(\sqrt{1+\frac{4}{k(n)^2}}\sqrt{1+\frac{8}{k(n)^2}} \right) \right) = \lim_{n \rightarrow \infty} \dfrac{f(n)}{g(n)} = \lim_{n \rightarrow \infty} \dfrac{df/dn}{dg/dn}\\ = \dfrac{- \dfrac1{\sqrt{3}} \frac1{k^2n^{2/3}}}{- \dfrac{2}{3n^{4/3}}} = \lim_{n \rightarrow \infty} \dfrac{\sqrt{3}}{2} \dfrac{n^{2/3}}{k^2}$$

Now as $n \rightarrow \infty$, $k(n) \sim \dfrac1{12}(108n+108n)^{1/3} = \dfrac1{12} \times 6n^{1/3} = \dfrac{n^{1/3}}{2}$. Hence, $$k^2(n) \sim \dfrac{n^{2/3}}{4}$$

Hence, we finally get that $$\lim\limits_{n \to \infty} \left( \frac{n^{1/3}}{2} \text{arcsec} \left(\sqrt{1+\frac{4}{k(n)^2}}\sqrt{1+\frac{8}{k(n)^2}} \right) \right) = \lim_{n \rightarrow \infty} \dfrac{\sqrt{3}}{2} \dfrac{n^{2/3}}{k^2} = \dfrac{\sqrt{3}}{2} \times 4 = 2 \sqrt{3}$$

Hence, the limit is $2 \sqrt{3}$. The approximate value is $3.464$ which is what you have mentioned in your comments (though the answer should be $2\sqrt{3}$ and not $3 \sqrt{2}$)


EDIT As Peter Tamaroff rightly points out in the comment, if we are looking at the limit of the sequence, then we need to apply the Stolz Cesaro rule (the equivalent of l'Hospital's rule for sequences). However, if the function converges to a limit, then so will the sequence, which is the case here. However, note that the converse is not true in general. For instance, look here for a counterexample!. However, it is easier to apply l'Hospital than Stolz Cesaro, and if we find that the limit exists from l'Hospital rule, we are fine.

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Why would you use L'Hopital's Rule for a sequence? Shouldn't you be using Stolz Cesaro, for example? –  Pedro Tamaroff May 24 '12 at 21:57
    
@PeterTamaroff Well... Aren't both the same? except for some minor changes. –  user17762 May 24 '12 at 21:59
    
Not really. You can argue that if the function goes to such limit, then the sequence will, which is the case here, but it should be noted, since the converse is not true. –  Pedro Tamaroff May 24 '12 at 22:01
    
@PeterTamaroff True. I accept this. But most often you can get away with this since it is easier to apply l'Hospital than Stolz Cesaro. –  user17762 May 24 '12 at 22:03
    
That's true, but you should make it explicit in the question, before applying L'Hôpital, or some people might take it for true in any case. –  Pedro Tamaroff May 24 '12 at 22:06

As $k$ is n-dependent first try to find limit $\lim\limits_{n \to \infty} (k_{n}-1) = \lim\limits_{n \to \infty}((\frac{1}{12} (108n+12 \sqrt{768+81n^2})^{1/3}-\frac{4}{ (108n+12 \sqrt{768+81n^2})^{1/3}}+1)-1)$ - it should make clear what is the limit of argument of $arccos$, next use fact that $\lim\limits_{n \to \infty}a_{n}b_{n}=\lim\limits_{n \to \infty}a_{n}\lim\limits_{n \to \infty}b_{n}$.

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$k \to \infty$ when $n\to \infty$. –  Sh.N. May 21 '12 at 19:52
    
$\lim\limits_{n \to \infty} a_{n}b_{n} \neq \lim\limits_{n \to \infty} a_{n}\lim\limits_{n \to \infty} b_{n}$. Let $a_{n}=1, b_{n}=1/n$. –  Sh.N. May 21 '12 at 19:53
    
Your first suggestion is correct and second is in particular not. Notice that in case of your $a_{n}b_{n}=b_{n}$. However in general case I should had added that $\lim\limits_{n \to \infty}a_{n}b_{n}=\lim\limits_{n \to \infty}a_{n}\lim\limits_{n \to \infty}b_{n}$ holds if both sequences converges. –  data May 21 '12 at 20:06
    
Going back to your limit, it is clear that as $n \rightarrow \infty $ arccos approaches $arccos(1)$ which is finite and $n^{1/3}\rightarrow \infty$ so I guess that given sequence diverges. –  data May 21 '12 at 20:12
    
We have uncertainty $0/0$ and, generally speaking, we should use L'Hopitals rule. I want to find another way. –  Sh.N. May 21 '12 at 21:12

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