# Find a value of $\;\lim\limits_{n\rightarrow\infty}n\left ( 1- a_{n} \right )$

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.

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

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

• Thanks a rl lot Mar 23, 2021 at 3:10