Computing the limit of $n \cdot \arccos \left( \left(\frac{n^2-1}{n^2+1}\right)^{\cos (1/n)} \right)$ 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.
 A: 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 $$2.$$
A: 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
$$\sqrt2\sqrt2=2$$
A: 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 $L = \sqrt{2} \sqrt{A} = 2$.
A: 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.
Then
$$\lim_{x\to0}\frac{\sqrt{2\left(1-\dfrac{1-x}{1+x}\right)}}{\sqrt x}=2.$$
