<s>[INST] Is this sequence of functions uniformly convergent on [0, 2] ?? Define a sequence of functions $f_n : [0,2] \to \Bbb R$ as:
$$f_n(x) = \frac {1-x} {1+x^n}$$
Is this sequence of functions uniformly convergent on $[0,2]$?
 [/INST] Let $f_n(x)=\frac{1-x}{1+x^n}$ and $f(x)=\begin{cases}1-x&,0\le x\le 1\\\\0&,1\le x\le 2\end{cases}$
Clearly we have
$$\lim_{n\to \infty}f_n(x)=f(x)$$
Furthermore, we see that 
$$|f_n(x)-f(x)|=\begin{cases}\frac{(1-x)x^n}{1+x^n}&,0\le x\le 1\\\\\frac{x-1}{1+x^n}&,1\le x\le 2\end{cases}$$

Next, we have the following estimates for $x\in [0,1]$
$$\begin{align}
\frac{(1-x)x^n}{1+x^n}&\le (1-x)x^n\\\\
&\le \left(\frac{1}{n+1}\right)\left(\frac{n}{n+1}\right)^n\\\\
&<\frac{1}{n+1}\\\\
&<\frac{1}{n-1}\\\\
&<\epsilon
\end{align}$$
whenever $n>1+\frac1\epsilon$.

Similarly, we have the following estimates for $x\in[1,2]$
$$\begin{align}
\frac{x-1}{1+x^n}&\le (x-1)x^{-n}\\\\
&\le \left(\frac{1}{n-1}\right)\left(\frac{n-1}{n}\right)^n\\\\
&<\frac{1}{n-1}\\\\
&<\epsilon
\end{align}$$
whenever $n>1+\frac1\epsilon$.

Putting it all together, we see that for all $\epsilon>0$ 
$$|f_n(x)-f(x)|<\epsilon$$
whenever $n>1+\frac1\epsilon$ for all $x\in [0,2]$.


The convergence is uniform.

</s>