Prove that $\{f_n\}$ converges pointwise but not uniformly on $\Bbb R$, where $f_n(x)=\frac{1-|x|^n}{1+|x|^n}$ . 
For each $n\in \Bbb N$ and $x\in \Bbb R$, define
$$\begin{align} f_n(x)=\frac{1-|x|^n}{1+|x|^n}_. \end{align} $$
Prove that $\{f_n\}$ converges pointwise but not uniformly on $\Bbb R$.

For $x\in (-1,1)$, $f_n(x)\to 1$, and for $x>1$ or $x< -1$, $ f_n(x)=\frac{1-|x|^n}{1+|x|^n} \to -1$. Define the limit function $\begin{align} f(x)=\begin{cases}
1,  & \text{if $x\in (-1,1)$ } \\
-1, & \text{if $x>1$ or $x< -1$}
\end{cases}_. \end{align}$ Then $f_n \to f$ pointwise on $\Bbb R$.
Let $\varepsilon_n=\sup\{\left|f(x)-f_n(x) \right|:x\in\Bbb R \}$, how can I get a contradiction by showing that $\varepsilon_n$ do not converge to $0$?

Your function $f$ is not complete: for $|x|=1$ we have $f_n(x) = 0 \to 0,$ hence $f(x)=0.$
The limit function $f$ is not continuous, but all $f_n$ are continuous ! Conclusion ?
