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.
Best Answer
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: $n(\sqrt[n]{L} - 1) \to \ln L, L > 0$