Calculate $\lim_{n\rightarrow\infty}\frac{\int_{0}^{1}f^n(x)\ln(x+2)dx}{\int_{0}^{1}f^n(x)dx}$ 
Given
$$f(x)=1-x^2+x^3 \qquad x\in[0,1]$$
calculate
$$
\lim_{n\rightarrow\infty}\frac{\int_{0}^{1}f^n(x)\ln(x+2)dx}{\int_{0}^{1}f^n(x)dx}
$$
where $f^n(x)=\underbrace{f(x)·f(x)·\dots\text{·}f(x)}_{n\ \text{times}}$.

This is a question from CMC(Mathematics competition of Chinese)in $2017$. The solution provides an idea: given  $s∈(0,\frac{1}{2}),$ prove:$$\lim_{n\rightarrow\infty}\frac{\int_{s}^{1}f^n(x)dx}{\int_{0}^{s}f^n(x)dx}=0\\$$ The final result is $\ln2.$

My approach
For this:$$\lim_{n\rightarrow\infty}\frac{\int_{s}^{1}f^n(x)dx}{\int_{0}^{s}f^n(x)dx}=0\\$$
I want to do piecewise calculation:$$\int_{s}^{1-s}f^n(x)dx+\int_{1-s}^{1}f^n(x)dx.$$For this:$$\lim_{n\rightarrow\infty}\frac{\int_{1-s}^{1}f^n(x)dx}{\int_{0}^{s}f^n(x)dx}=0.\\$$Here is the proof: when$\ \ n≥\frac{1}{s^2}$, $$\frac{\int_{1-s}^{1}f^n(x)dx}{\int_{0}^{s}f^n(x)dx}=\frac{\int_{0}^{s}(1-x^2(1-x))^ndx}{\int_{0}^{s}(1-x(1-x)^2)^ndx}\\\leq\frac{\int_{0}^{s}(1-\frac{x}{4})^ndx}{\int_{0}^{s}(1-x^2)^ndx}\leq\frac{\int_{0}^{s}(1-\frac{x}{4})^ndx}{\int_{0}^{1/\sqrt{n}}(1-\frac{x}{\sqrt{n}})^ndx}\\=\frac{\frac{4}{n+1}(1-(1-\frac{s}{4})^{n+1})}{\frac{\sqrt{n}}{n+1}(1-(1-\frac{1}{n})^{n+1})}\sim\frac{4}{\sqrt{n}(1-\frac{1}{e})}\rightarrow0.\\$$For this:$$\lim_{n\rightarrow\infty}\frac{\int_{s}^{1-s}f^n(x)dx}{\int_{0}^{s}f^n(x)dx}=0.\\$$Here is the proof: given $t,0<t<s<\frac{1}{2},$then$$f(t)>f(s)>f(1-s).$$Define $m_t=\min_{x\in[0,t]}f(x),M_s=\max_{x\in[s,1-s]}f(x),$ so$$m_t=f(t)>f(1-s)=M_s.$$$$\frac{\int_{s}^{1-s}f^n(x)dx}{\int_{0}^{s}f^n(x)dx}\leq\frac{\int_{s}^{1-s}f^n(x)dx}{\int_{0}^{t}f^n(x)dx}$$$$\leq\frac{(1-2s)M_s ^n}{tm_t ^n}=\frac{1-2s}{t}(\frac{M_s}{m_t})^n\rightarrow0.\\$$ In conclusion,we can get:$$\lim_{n\rightarrow\infty}\frac{\int_{s}^{1}f^n(x)dx}{\int_{0}^{s}f^n(x)dx}=0.$$
 A: Fisrtly,we prove that:$$\lim_{n\rightarrow\infty}\frac{\int_{s}^{1}f^n(x)dx}{\int_{0}^{s}f^n(x)dx}=0\\$$
Perform piecewise calculation:$$\int_{s}^{1-s}f^n(x)dx+\int_{1-s}^{1}f^n(x)dx.$$For this:$$\lim_{n\rightarrow\infty}\frac{\int_{1-s}^{1}f^n(x)dx}{\int_{0}^{s}f^n(x)dx}=0.\\$$When$\ \ n≥\frac{1}{s^2}$, $$\frac{\int_{1-s}^{1}f^n(x)dx}{\int_{0}^{s}f^n(x)dx}=\frac{\int_{0}^{s}(1-x^2(1-x))^ndx}{\int_{0}^{s}(1-x(1-x)^2)^ndx}\\\leq\frac{\int_{0}^{s}(1-\frac{x}{4})^ndx}{\int_{0}^{s}(1-x^2)^ndx}\leq\frac{\int_{0}^{s}(1-\frac{x}{4})^ndx}{\int_{0}^{1/\sqrt{n}}(1-\frac{x}{\sqrt{n}})^ndx}\\=\frac{\frac{4}{n+1}(1-(1-\frac{s}{4})^{n+1})}{\frac{\sqrt{n}}{n+1}(1-(1-\frac{1}{n})^{n+1})}\sim\frac{4}{\sqrt{n}(1-\frac{1}{e})}\rightarrow0.\\$$For this:$$\lim_{n\rightarrow\infty}\frac{\int_{s}^{1-s}f^n(x)dx}{\int_{0}^{s}f^n(x)dx}=0.\\$$Given $t,0<t<s<\frac{1}{2},$then$$f(t)>f(s)>f(1-s).$$Define $m_t=\min_{x\in[0,t]}f(x),M_s=\max_{x\in[s,1-s]}f(x),$ so$$m_t=f(t)>f(1-s)=M_s.$$$$\frac{\int_{s}^{1-s}f^n(x)dx}{\int_{0}^{s}f^n(x)dx}\leq\frac{\int_{s}^{1-s}f^n(x)dx}{\int_{0}^{t}f^n(x)dx}$$$$\leq\frac{(1-2s)M_s ^n}{tm_t ^n}=\frac{1-2s}{t}(\frac{M_s}{m_t})^n\rightarrow0.\\$$ In conclusion,we can get:$$\lim_{n\rightarrow\infty}\frac{\int_{s}^{1}f^n(x)dx}{\int_{0}^{s}f^n(x)dx}=0.$$
Secondly,we calculate the result:$\ln2$.
For $\varepsilon\in(0,\ln\frac{5}{4})$,take $s=2(e^\varepsilon-1)$,than $s\in(0,\frac{1}{2}),\ln\frac{2+s}{2}=\varepsilon.$
From the above conclusion,we can know:
$\exists N\geq1,s.t.$,when $n\geq N$,$$\frac{\int_{s}^{1}f^n(x)dx}{\int_{0}^{s}f^n(x)dx}\leq \varepsilon.$$So$$\lvert\frac{\int_{0}^{1}f^n(x)\ln(x+2)dx}{\int_{0}^{1}f^n(x)dx}-\ln2\rvert=\frac{\int_{0}^{1}f^n(x)\ln\frac{x+2}{2}dx}{\int_{0}^{1}f^n(x)dx}$$$$\leq\frac{\int_{0}^{s}f^n(x)\ln\frac{x+2}{2}dx}{\int_{0}^{s}f^n(x)dx}+\frac{\int_{s}^{1}f^n(x)\ln\frac{x+2}{2}dx}{\int_{0}^{s}f^n(x)dx}$$$$\le\ln\frac{s+2}{2}+\frac{\ln\frac{3}{2}\int_{s}^{1}f^n(x)dx}{\int_{0}^{s}f^n(x)dx}\leq\varepsilon(1+\ln\frac{3}{2}).\\$$Let $\varepsilon\rightarrow0$,we can get$$\lim_{n\rightarrow\infty}\frac{\int_{0}^{1}f^n(x)\ln(x+2)dx}{\int_{0}^{1}f^n(x)dx}=\ln2.\\$$That's all.
