TITLE: Find a value of $\;\lim\limits_{n\rightarrow\infty}n\left ( 1- a_{n} \right )$
QUESTION [9 upvotes]: I'm going to give you an extremal problem.
Given $a_{n}:=\left ( \text{the solution for the equation}\,x^{n}= \cos x,\quad x> 0 \right ),$ find a value of
$$\lim_{n\rightarrow\infty}n\left ( 1- a_{n} \right )$$
Source: Fujino_Yusui

If you imagine the graph of $x^{n},$ it should intersect at the point where it increases rapidly by $1,$ which is about where $x^{n}= \cos 1\,(y$ is increasing rapidly, so if you look at $y,$ you should be able to approximate $x$ well enough$).\,n\left ( 1- \cos^{\frac{1}{n}}1 \right )$ looks good, and $\frac{1}{n}= h$ is the derivative of $h\rightarrow 0^{+}.$ I guess $-\ln\cos 1$ by definition.

REPLY [5 votes]: Here is an easy solution.
$x^n=\cos x$,  as $a_n$ satisfies the equation, $\left(a_n\right)^n=\cos \left(a_n\right)$
Taking log on both sides,
$n.\log \left(a_n\right)=\log \left(\cos \left(a_n\right)\right)$
$\therefore n=\frac{\log \left(\cos \left(a_n\right)\right)}{\log \left(a_n\right)}$
Now as n tends to infinity, the rhs must also tend to infinity. As cos lies between 0-1, only alternative is that $a_n$ tends to 1.  ($a_n=\pi/2$ wil never satisfy the original equation)
$n\rightarrow \infty \therefore a_n\rightarrow 1$
Now we substitute this value of n in our limit,
$\lim _{n\rightarrow \infty }\left(n\left(1-a_n\right)\right)=\lim _{a_n\rightarrow 1}\left(\frac{\log \left(\cos \left(a_n\right)\right)}{\log \left(a_n\right)}\left(1-a_n\right)\right)$
As it is now a 0/0 form, we use the l's hospital, and get the required answer as $a_\infty=-\ln\cos 1$