Find $\lim\limits_{n\to+\infty}n(1-a_n)$ Where $a_n>0$ satisfies $\cos a_n=a_n^n$ Suppose the sequence $1\ge\{a_n\}>0$ is defined by one of the positive root of $$\cos x=x^n$$
How to find the limit:
$$\lim_{n\to+\infty}n(1-a_n)$$
with the increase of $n$, it seems, the limit of $a_n$ is 1 from the observation below:

 A: Note that $a_n<1$, hence
$$a_n=\sqrt[n]{\cos(a_n)}>\sqrt[n]{\cos(1)}$$
From this, we may also conclude that
$$a_n=\sqrt[n]{\cos(a_n)}<\sqrt[n]{\cos\left(\sqrt[n]{\cos(1)}\right)}$$
But,
$$\lim_{n\to\infty}n\left(1-\sqrt[n]{\cos(1)}\right)=-\ln(\cos(1))$$
$$\lim_{n\to\infty}n\left(1-\sqrt[n]{\cos\left(\sqrt[n]{\cos(1)}\right)}\right)=-\ln(\cos(1))$$
Both of which may be derived by rewriting the limits as
$$\lim_{n\to\infty}n(1-f(1/n))=\lim_{x\to0}\frac{f(0)-f(x)}x=-f'(0) $$
A: You seem to already no that $f_n(x):=x^n-\cos(x)$ has only one positive root; we see this by observing that $f_n'(x)>0$ for $x\in(0,\pi]$. The key point is to now notice the following: For $\delta>0$ we have
$$
\lim_{n\to\infty}f_n\left(1+\frac{\log(\cos(1))\pm\delta}{n}\right)=\cos(1)\left(e^{\pm\delta}-1\right).
$$
As $\delta>0$ we observe that $\cos(1)\left(e^{-\delta}-1\right)<0<\cos(1)\left(e^{\delta}-1\right)$. Thus there exists an $N_\delta\in\mathbb{N}$ such that for all $n\geq N_\delta$
$$
f_n\left(1+\frac{\log(\cos(1))-\delta}{n}\right)<0<f_n\left(1+\frac{\log(\cos(1))+\delta}{n}\right).
$$
But as $f_n$ is strictly increasing, we thus have
$$
1+\frac{\log(\cos(1))-\delta}{n}<a_n<1+\frac{\log(\cos(1))+\delta}{n}\qquad\forall n\geq N_\delta.
$$
Can you take from here?
