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.

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    $\begingroup$ why the tags recursion and recurrence-relations ? $\endgroup$
    – René Gy
    Commented Mar 21, 2021 at 8:29
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    $\begingroup$ It is my habit. $\endgroup$
    – user822157
    Commented Mar 21, 2021 at 8:31
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    $\begingroup$ @HanulJeon, thanks for your edit, I often don't link the Twitter's account with my post. $\endgroup$
    – user822157
    Commented Mar 21, 2021 at 9:29

2 Answers 2


Look at the function $f(x) = x^n - \cos x\implies f'(x) = nx^{n-1}+\sin x > 0$ on $(0,1)$ and $f(0)\cdot f(1) < 0 \implies \exists ! a_n\in (0,1): f(a_n) = 0\implies 0 < a_n < 1$. Next you show $\{a_n\}$ is an increasing sequence ( I leave this for OP to tackle : it will be a good wrestling match for ya ) and you have to use the $x^n = \cos x$ to do it !. After that then it has a limit $L$ and $L = \left(\cos L\right)^{0} = 1$. Thus: $n(1-a_n)= -n(-1+\sqrt[n]{\cos a_n})\to -\ln(\cos 1)$. Note that you can show: $n(\sqrt[n]{L} - 1) \to \ln L, L > 0$


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 $a_\infty=-\ln\cos 1$

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    $\begingroup$ Thanks a rl lot $\endgroup$
    – user822157
    Commented Mar 23, 2021 at 3:10

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