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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?2012-05-21
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    @Eepzy No, I haven't.2012-05-21
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    I tried to use L'Hopital's rule, but expressions are enormous.2012-05-21
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    I computated the function under 'lim' for large $n$. It is approximately $3\sqrt{2}$ (or $3.4$).2012-05-21
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    @Sh.N. Could you tell us the source of this problem?2012-05-24

2 Answers 2

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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?2012-05-24
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    @PeterTamaroff Well... Aren't both the same? except for some minor changes.2012-05-24
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    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.2012-05-24
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    @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.2012-05-24
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    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.2012-05-24
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    @PeterTamaroff Yes I will but what makes you think that the OP is talking about a sequence and not a function? :)2012-05-24
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    $n$ variable. It is convention to use $n$ for natural numbers and speacially for sequences. Do you agree?2012-05-24
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    @PeterTamaroff Yes. I just wanted to bring to notice that how we always think that $n$ stands for natural numbers, where $x$ stands for reals even when nothing is explicitly mentioned :).2012-05-24
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    Rightfully upvoted now.2012-05-25
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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 $n\to \infty$.2012-05-21
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    $\lim\limits_{n \to \infty} a_{n}b_{n} \neq \lim\limits_{n \to \infty} a_{n}\lim\limits_{n \to \infty} b_{n}$. Let $a_{n}=1, b_{n}=1/n$.2012-05-21
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    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.2012-05-21
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    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.2012-05-21
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    We have uncertainty $0/0$ and, generally speaking, we should use L'Hopitals rule. I want to find another way.2012-05-21