Does $f_n(x)=\frac{1}{nx}$ converge uniformly for $x \in ]0,1]$? Does $(f_n)_{n=1}^{\infty}$, $f_n(x)=\frac{1}{nx}$ converge uniformly for $x \in ]0,1]$?
I'd say yes, and actually for all $x \in \mathbb{R}$, because the above is equal to
$$\frac{1}{x}\lim_{n \rightarrow \infty} \frac{1}{n}=\frac{1}{x} \cdot 0 = 0 \text{ when } n \rightarrow \infty$$
However, my exercise is posited so as if there was some problem in the area $x\in [1/2,1]$. But how could it be?
 A: First, notice that the functions $f_n$ are not defined on $0$ so there is no sense to say that the uniform convergence is actually on all $\Bbb R$. Secondly the point-wise convergence of $f_n$ is the null function on $(0,1]$ and we have
$$\Vert f_n\Vert_\infty=\sup_{x\in(0,1]}\vert f_n(x)\vert\ge f_n\left(\frac1n\right)=1\not\xrightarrow{n\to\infty}0$$
so the convergence is not uniform on $(0,1]$.
A: 
Recall that the definition of the negation of uniform convergence states that a sequence of functions $f_n(x)$ , which converges to $f(x)$, fails to converge uniformly to $f(x)$ for $x\in A$ if there exists an $\epsilon>0$, such that for all $N\in \mathbb{N}$, there exists an $x\in A$ and a number $n>N$ such that
$$|f_n(x)-f(x)|\ge \epsilon$$

If $f_n(x)=\frac1{nx}$, $f(x)=0$, and $A=\{x|\,0<x\le 1\}$, then for $\epsilon =1/2$ and all $N$, there exists an $x=1/N\in (0,1]$ and a number $n=2N>N$ such that 
$$\begin{align}
\left|f_n(x)-0\right|&=\left|\frac{1}{nx}-0\right|\\\\
&=\left|\frac{N}{n}\right|\\\\
&=\frac12\\\\
&\ge \epsilon
\end{align}$$

There, we conclude that $f_n(x)$ fails to converge uniformly to $0$ for $x\in (0,1]$.



Recall that the definition of uniform convergence states that a sequence of functions $f_n(x)$ , which converges to $f(x)$, converges uniformly to $f(x)$ for $x\in B$ if for all $\epsilon>0$, there exist a number $N\in \mathbb{N}$, such that for all $x\in B$ and all $n>N$ 
$$|f_n(x)-f(x)|< \epsilon$$

Let $B=\{x\,|\,\delta \le x\le 1\}$, for $\delta>0$ and let $\epsilon>0$ be given.  Then, note that 
$$\left|f_n(x)-0\right|=\frac{1}{nx}\le \frac{1}{n\delta }<\epsilon$$
for all $x\in B$ and for all $n>\frac{1}{\delta \epsilon}$. 
Hence, $f_n(x)$ converges uniformly to $0$ for $x\in [\delta,1]$ for $\delta >0$.
A: The farther you get x towards zero, the bigger 1/x gets and therefore you need to choose a bigger n to get $\frac{1}{nx}<\epsilon$. Therefore the convergence can't be uniform.
On any compact interval like [0.5,1] there should not be a problem, because 1/x will have a maximum value on the interval, so if you take n with $\frac{max\frac{1}{x}}{n}<\epsilon $ This n will work for every x in the interval.
