Let $C[0,1]$ be the vector space of continuous real-valued functions defined on $J:=[0,1]$. It can be made into a normed space if one sets $$||f(x)||=\max_{x\in J}{|f(x)|}.$$ I'm trying to show that $\lim_\limits{n\to \infty} ||f_n-F||=0$, where $$f_n(x)=-|2x-1|^{1+1/n}+1 \ \ \ \mbox{and} \ \ \ F(x)=-|2x-1|+1.$$ So, for all $\epsilon>0$, I need to find $N>0$ such that $$n>N \Rightarrow \max_{x\in J}\left\{|2x-1|-|2x-1|^{1+1/n} \right\}<\epsilon,$$ or $$n>N \Rightarrow \max_{x\in J}\left\{|2x-1|\left[1-|2x-1|^{1/n}\right]\right\}<\epsilon.$$
My attempt. I guess I have to use $\max_{x\in J}{|2x-1|}=1$ and find a expression $B(n)$ which depends on $n$ but not on $x$ such that $1-|2x-1|^{1/n}<B(n)<\epsilon$ and $B(n)\rightarrow0$. But I couldn't find that. Then, I'd like someone to give a hint, please. Thanks in advance!
Since $|2x-1|$ is symmetric to $x=\frac12$, we only discuss the case of $2x-1\ge0$. Note $0\le2x-1\le1$ for $x\in[1/2,1]$. Let \begin{eqnarray} g_n(x)&=&(2x-1)-(2x-1)^{1+1/n}. \end{eqnarray} Clearly $g_n(x)\ge 0$ for $x\in[1/2,1]$. Leting $g_n'(x)=0$ gives $x=x_n\equiv\frac12\left[1+\left(\frac{n}{n+1}\right)^n\right]$ which is $[1/2,1]$. Note $$ g_n''(x_n)=-\frac4n(1+\frac1n)^n<0$$ namely $g_n$ obtains the max at $x=x_n$. Hence $$ \|f_n-F\|=\max_{x\in[1/2,1]}g_n(x)=g_n(x_n)=\frac1{n+1}(\frac{n}{n+1})^n$$ and thence $$ \lim_{n\to\infty}\|f_n-F\|=\lim_{n\to\infty}\frac1{n+1}(\frac{n}{n+1})^n=0. $$