Pointwise convergence of uniformly continuous functions to zero, but not uniformly What would be an example of a sequence of uniformly continuous functions on a compact domain which converges pointwise to $0$, but not uniformly? 
 A: The another standard one is the growing steeple on $[0,1]$:
$$f_n(x)=\begin{cases}n^2 x &\text{if}\;0 \leq x \leq \frac{1}{n}\\ 2n-n^2 x &\text{if}\;\frac{1}{n} \leq x \leq \frac{2}{n}\\ 0 &\text{if}\;\frac{2}{n} \leq x \leq 1 \end{cases}$$
Then each $f_n$ is uniformly continuous and also the limit is zero, but the convergence is not uniform.
Added: It is easy to visualize the graph of $f_n$. Actually each $f_n$ is a triangle with height  $n$ attained at $1/n$ 
A: Consider 
$$f_n(x)=1-\min(1,n|x-1/n|)=\begin{cases}
nx&\text{ if }x<\frac{1}{n}\\2-nx&\text{ if }\frac1{n}\leq x\leq\frac{2}{n}\\0&\text{ otherwise.}\end{cases}$$
Each $f_n$ is continuous on the compact set $[0,1]$ and therefore it is also uniformly continuous. Moreover,  $f_n(x)\to 0$ for any $x\in [0,1]$, but the convergence is not uniform on $[0,1]$ because $\max_{x\in[0,1]}|f_n(x)|=f(1/n))=1$.
A: Take$$\begin{array}{rccc}f_n\colon&[0,1]&\longrightarrow&\mathbb R\\&x&\mapsto&nx^n(1-x).\end{array}$$The sequence $(f_n)_{n\in\mathbb N}$ converges pointwise to the null function, but not uniformly, since$$(\forall n\in\mathbb N):f_n\left(\frac n{n+1}\right)=\left(\frac n{n+1}\right)^{n+1}$$and $\lim_{n\to\infty}\left(\frac n{n+1}\right)^{n+1}=e^{-1}$.
