I need to solve this earth's wonder:

$$\lim_{n \rightarrow \infty} \left[n \; \arccos \left( \left(\frac{n^2-1}{n^2+1}\right)^{\cos \frac{1}{n}} \right)\right]$$

I have tried to write down it using $e^{v \ln u}$,and then used L'Hôpital's rule, but with no luck, i'm constantly getting indeterminate form like $\infty-\infty$ inside $\ln$ which makes another application of L'Hôpital impossible.

My professor told me (with great smile on its face) that if i use Taylor expansion, it will lead me into the abyss...

any hints about possible rewriting this limit and what i should use would be VERY helpful. Thanks in advance.

  • 1
    $\begingroup$ Please, use $\large\left(\phantom{ABC}\right)$'s to clarify your question. $\endgroup$ Feb 4, 2014 at 20:35
  • $\begingroup$ To write parentheses in LaTeX (/MathJax), use ( ). The brackets { } are just used to group expressions internally. $\endgroup$ Feb 4, 2014 at 20:37
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    $\begingroup$ @user3116147 It's not obvious. In fact it's still unclear--do you apply $\arccos$ before taking the exponent, or do you first apply $\arccos$ and then take the result to the power of $\cos \frac1n$? $\endgroup$ Feb 4, 2014 at 20:46
  • $\begingroup$ it should have been $arccos$ of whole expression that follow $\endgroup$
    – Zoran
    Feb 4, 2014 at 20:51

4 Answers 4


This answer assumes that one is looking for $$\lim_{n \to\infty} n \cdot\arccos \left[\left(\frac{n^2-1}{n^2+1}\right)^{\cos(1/n)} \right]. $$

Quote: My professor told me (with (a) great smile on (their) face) that if (I) use Taylor expansion(s), it will lead me into the abyss...

To the abyss then! :-) To begin with, note that $$(n^2-1)/(n^2+1)=1-2/n^2+o(1/n^2)$$ and that $\cos(1/n)\to1$ hence the argument of the arccos is $$1-2/n^2+o(1/n^2).$$ Furthermore, $\cos x=1-x^2/2+o(x^2)$ when $x\rightarrow 0$ hence $$\arccos(1-x^2/2)=|x|+o(x)$$ when $x\to0$. Applying this to $x^2/2=2/n^2+o(1/n^2)$, one sees that the arccos is $2/n+o(1/n)$, hence the whole limit is $$2.$$

  • $\begingroup$ Haha...nice...thank you very,very much $\endgroup$
    – Zoran
    Feb 4, 2014 at 20:58
  • $\begingroup$ yes,it looks just as it should be....thank you one more time $\endgroup$
    – Zoran
    Feb 4, 2014 at 21:00
  • $\begingroup$ @mischek The edit you suggested was wrong. Users Yiorgos S. Smyrlis and Matthew Conroy: Please be more careful before accepting mathematically inaccurate edits. $\endgroup$
    – Did
    Feb 4, 2014 at 22:12
  • $\begingroup$ Oh my, I really do need to study Taylor series properly. $\endgroup$
    – Dahn
    Feb 4, 2014 at 22:25
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    $\begingroup$ @mischek Yes, but $\frac{x^4}{4!} - \frac{x^6}{6!} + \cdots$ is $o(x^2)$--and not $o(x^4)$--as $x \to 0$. $\endgroup$ Feb 5, 2014 at 4:08

First, rewriting things with $u=1/n$, the problem becomes

$$\lim_{u\to0}{\arccos\left(\left({1-u^2\over1+u^2}\right)^{\cos u}\right)\over u}$$

This can be rewritten as

$$\lim_{u\to0}\left({\arccos\left(\left({1-u^2\over1+u^2}\right)^{\cos u}\right)\over \sqrt{1-\left({1-u^2\over1+u^2}\right)^{\cos u}} } \right) \sqrt{\lim_{u\to0}\left({1-\left({1-u^2\over1+u^2}\right)^{\cos u}\over u^2}\right)} $$

provided these two limits exist (which is what we're about to show). For the first, let

$$x=\left({1-u^2\over1+u^2}\right)^{\cos u}$$

note that the limit as $u\to0$ becomes

$$\lim_{x\to1}{\arccos x\over\sqrt{1-x}}=\lim_{x\to1}{-1/\sqrt{1-x^2}\over-1/(2\sqrt{1-x})}=\lim_{x\to1}{2\over\sqrt{1+x}}=\sqrt2$$

using L'Hopital and then simplifying. For the second, L'Hopital gives us

$$\lim_{u\to0}{1-e^{\cos u(\ln(1-u^2)-\ln(1+u^2)}\over u^2}=\lim_{u\to0}{\sin u(\ln(1-u^2)-\ln(1+u^2))-\cos u\left({-2u\over1-u^2}-{2u\over1+u^2} \right)\over 2u}\left({1-u^2\over1+u^2}\right)^{\cos u} =\lim_{u\to0}\left({1\over2}{\sin u\over u}\ln\left({1-u^2\over1+u^2}\right)+\cos u\left({1\over1-u^2}+{1\over1+u^2} \right)\right)\left({1-u^2\over1+u^2}\right)^{\cos u}=\left({1\over2}(1)(\ln1)+(1)(1+1)\right)(1)^1=2$$

Thus both limits, as promised, exist, and combined (remembering to take the square root of the second limit) we get the limit



Very informally:

Let $x:=1/n^2$.

By Taylor, for small $x$ (i.e. $y$ close to $1$), $y=\cos x\approx1-x^2/2$ so that $x=\arccos y\approx\sqrt{2(1-y)}$.

Then the argument of the $\arccos$ is of the form $((1-x)/(1+x))^{1-x/2+\cdots}$. The exponent can be simplified to $1$ because by the generalized binomial theorem $(1+\epsilon)^{1+\eta}\approx1+(1+\eta)\epsilon\approx1+\epsilon$ and higher order terms.


$$\lim_{x\to0}\frac{\sqrt{2\left(1-\dfrac{1-x}{1+x}\right)}}{\sqrt x}=2.$$


The function given here is really bit complex (at least in typing). We first need to check if we know any fundamental limit associated with $\arccos$ function. Clearly there isn't any, but we do have the standard limit $$\lim_{x \to 0}\frac{1 - \cos x}{x^{2}} = \lim_{x \to 0}\dfrac{2\sin^{2}(x/2)}{x^{2}} = \frac{1}{2}$$ Putting $\cos x = t$ we can easily see that $$\lim_{t \to 1^{-}}\dfrac{\arccos^{2}t}{1 - t} = 2$$ and further putting $1 - t = x$ we get $$\lim_{x \to 0^{+}}\dfrac{\arccos^{2}(1 - x)}{x} = 2$$ or $$\lim_{x \to 0^{+}}\frac{\arccos(1 - x)}{\sqrt{x}} = \sqrt{2}\tag{1}$$ Next we focus on the complicated function given in the question. If we put $x = 1/n$ then we can see that the function is given by $$f(x) = \frac{1}{x}\cdot\arccos\left(\left(\frac{1 - x^{2}}{1 + x^{2}}\right)^{\cos x}\right)$$ If we put $y = ((1 - x^{2})/(1 + x^{2}))^{\cos x}$ then $y \to 1^{-}$ and hence $z = 1 - y \to 0^{+}$. And we have \begin{align} L &= \lim_{x \to 0^{+}}f(x)\notag\\ &= \lim_{x \to 0^{+}}\frac{1}{x}\cdot\arccos(1 - z)\notag\\ &= \lim_{x \to 0^{+}}\frac{1}{x}\cdot\sqrt{z}\cdot\frac{\arccos(1 - z)}{\sqrt{z}}\notag\\ &= \sqrt{2}\lim_{x \to 0^{+}}\sqrt{\dfrac{z}{x^{2}}}\text{ (using (1))}\\ &= \sqrt{2}\lim_{x \to 0^{+}}\sqrt{\dfrac{1 - y}{x^{2}}}\tag{2} \end{align} Next we can see that \begin{align} 1 - y &= 1 - \left(\frac{1 - x^{2}}{1 + x^{2}}\right)^{\cos x}\notag\\ &= 1 - \left(1 - \frac{2x^{2}}{1 + x^{2}}\right)^{\cos x}\notag\\ &\geq 1 - \left(1 - \frac{2x^{2}\cos x}{1 + x^{2}}\right)\notag\\ &= \frac{2x^{2}\cos x}{1 + x^{2}}\tag{3} \end{align} and \begin{align} 1 - y &= 1 - \left(\frac{1 - x^{2}}{1 + x^{2}}\right)^{\cos x}\notag\\ &= 1 - \left(1 - \frac{2x^{2}}{1 + x^{2}}\right)^{\cos x}\notag\\ &\leq 1 - (1 - 2x^{2})^{\cos x}\notag\\ &\leq 1 - (1 - 2x^{2})\notag\\ &= 2x^{2}\tag{4} \end{align} It follows from $(3)$ and $(4)$ that $$\frac{2\cos x}{1 + x^{2}}\leq \frac{1 - y}{x^{2}} \leq 2$$ and taking limits as $x \to 0^{+}$ and using Squeeze theorem we get $$\lim_{x \to 0^{+}}\frac{1 - y}{x^{2}} = 2$$ Using this limit in equation $(2)$ we get the desired limit $L = \sqrt{2}\cdot\sqrt{2} = 2$.

The above solution avoids the use of Taylor's series and instead relies on fundamental limit theorems. Using Taylor's series would be a challenge if we proceed directly. Did does it so smartly in his answer and I am at loss of words to appreciate his technique. So whenever we are asked to use Taylor's we must make certain optimizations (regarding no of terms of the series to be used) and this requires bit of experience to get the right answer in the most efficient manner. On the other hand avoiding Taylor's series or LHR almost always requires various algebraical/trigonometric manipulations and some ingenious use of inequalities and the use of Squeeze theorem (like I have done above).

Update: The limit of $(1 - y)/x^{2}$ can also be evaluated by expressing $y$ as $y = \exp(\log y)$ and this appears to be simpler than the approach via Squeeze theorem presented above. We have \begin{align} A &= \lim_{x \to 0^{+}}\frac{1 - y}{x^{2}}\notag\\ &= \lim_{x \to 0^{+}}\frac{1 - \exp(\log y)}{\log y}\cdot\frac{\log y}{x^{2}}\notag\\ &= -\lim_{x \to 0^{+}}\frac{\cos x\log((1 - x^{2})/(1 + x^{2}))}{x^{2}}\notag\\ &= \lim_{x \to 0^{+}}\frac{\log((1 + x^{2})/(1 - x^{2}))}{x^{2}}\notag\\ &= \lim_{x \to 0^{+}}\frac{\log(1 + x^{2})}{x^{2}} - \frac{\log(1 - x^{2})}{x^{2}}\notag\\ &= 1 - (-1) = 2\notag \end{align} and thus $L = \sqrt{2} \sqrt{A} = 2$.

  • $\begingroup$ @Did yes,i totally agree both with you and Did,and if i told you that my professor is mad enough to give this kind of problems just for passing the exam,you would agree that shooting myself is maybe the best option :)))...i dont even have the courage to post the problems for higher grade...thanks for your reply $\endgroup$
    – Zoran
    Feb 5, 2014 at 16:21

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