I want to determine the uniform convergence of the family of functions $ f_n:[0,2]\to\mathbb{R}$ defined by $ $ f_n(x)=\frac{1-x}{1+x^n}. $ $

**My attempt**

Let $ f$ be the limit of $ f_n$ , then $ $ f(x)=\begin{cases} 1-x,& 0\le x <1\ 0,&1\le x\le 2 \end{cases}. $ $ $ f$ is continuous on $ [0,1]$ , but not differentiable at $ x=1$ .

- Inequalities about $ |f_n(x)-f(x)|$

We know that $ $ |f_n(x)-f(x)| = \begin{cases} \frac{(1-x)x^n}{1+x^n},&0\le x<1\ \frac{x-1}{1+x^n},&1\le x\le 2 \end{cases}. $ $ If $ 0\le x<1$ , $ $ |f_n(x)-f(x)|=\frac{(1-x)x^n}{1+x^n} \le (1-x)x^n, $ $ and $ (1-x)x^n$ has the maximum $ \frac{1}{n}\left(1-\frac{1}{n+1}\right)^{n+1}$ at $ x=\frac{n}{n+1}$ . Also, if $ 1\le x\le 2$ , then $ $ |f_n(x)-f(x)|=\frac{x-1}{1+x^n}\le \frac{x-1}{x^n} $ $ and $ \frac{x-1}{x^n}$ has the maximum $ \frac{n-2}{(n-1)^2}\left(1-\frac{1}{n-1}\right)^{n-1}$ at $ x=\frac{n}{n-1}$ . Thus we obtain \begin{align} |f_n(x)-f(x)| &\le \max\left\{\frac{1}{n}\left(1-\frac{1}{n+1}\right)^{n+1},\frac{n-2}{(n-1)^2}\left(1-\frac{1}{n-1}\right)^{n-1}\right\}\ &\le \frac{1}{n-1} \end{align} for sufficiently large $ n$ .

For $ \epsilon > 0$ , there is $ N\in\mathbb{N}$ such that $ \frac{1}{N}<\epsilon$ by Archimedian property. Then, for any $ n\in\mathbb{N}$ and $ x\in [0,2]$ , if $ n > N$ , then $ $ |f_n(x)-f(x)|\le \frac{1}{n-1}\le \frac{1}{N} <\epsilon. $ $ Therefore $ f_n$ uniformly converges on $ [0,2]$ .

**“Solution”** $ f_n$ does not converge uniformly, since $ f_n$ is differentiable but $ f$ is not differentiable at $ x=1$ .

However, I think the solution is wrong. For example, $ f_n(x)=\frac{\sin nx}{\sqrt{n}}$ converges uniformly on $ \mathbb{R}$ , whereas $ f_n'(x)=\sqrt{n}\cos nx$ does not pointwisely converge.

**Question** Is my attempt correct? If so, is there an easier method to solve the problem?