Prove that $f_n(x)=\frac{x}{n}$, $n=1,2,\ldots$ does not converge uniformly on $\mathbb{R}$ 
Prove that $f_n(x)=\frac{x}{n}$, $n=1,2,\ldots$ does not converge uniformly on
  $\mathbb{R}$.

It's clear this function converges pointwise to $0$ function. We have to show that there is $\epsilon>0$ such that for any $N\in \mathbb{N}$ there are $n>N$ and $x$ such that $|\frac{x}{n}|\ge \epsilon$. Choose $\epsilon=1$. However big $N$ is, we can choose $x>N+1$ so we have $|x|\ge n$ if $n=N+1$.
Is this correct?
 A: Definition: We say a sequence of functions $\left\{ f_n \right\}$, $n = 1,2,3,\dots$ converges uniformly on a set $S$ to a function $f$ if for every $\epsilon > 0$ there is an integer $N$ such that $n \geq N$ implies $$|f_n - f| < \epsilon$$ for all $x \in S$.
Solution: As you point out, the sequence of functions $f_n = \displaystyle \frac{x}{n}$, $n=1,2,3,\dots$ converges pointwise to $0$ on $\mathbb{R}$. So we are trying to show that $f_n$ does not converge uniformly to $f = 0$ on $\mathbb{R}$.
You have the right idea for approaching the problem. According to the definition, to show that $\displaystyle f_n = \frac{x}{n}$ does not converge uniformly on $\mathbb{R}$, we must only find a single $\epsilon > 0$ for which the condition:
" there is an integer $N$ such that $n \geq N\implies |f_n - f| < \epsilon$ for all $x \in \mathbb{R}$ "
does not hold.  
As you suggest, let us take $\epsilon = 1$ (note our choice of $\epsilon$ is arbitrary) and suppose that $f_n$ does converge uniformly. If $f_n$ converges uniformly, then by the definition, there is an integer $N$ such that $n \geq N$ implies $$\displaystyle \left| \frac{x}{n} \right| < 1$$ for all $x \in \mathbb{R}$. This would mean that, for example, $$\displaystyle \left| \frac{x}{N} \right| < 1$$ for any $x \in \mathbb{R}$. But this is a contradiction since for any $x \geq N$ $$ \displaystyle \left| \frac{x}{N} \right| \geq 1 $$
Therefore $f_n = \frac{x}{n}$ cannot be uniformly convergent on $\mathbb{R}$.
The most crucial part of the definition is the order in which we are allowed to fix $N$ and $x$. For pointwise convergence, the order is reversed: we fix $x$ and then we are allowed to find an $N$ which satisfies the $\epsilon$ condition. For uniform convergence however, the definition requires that we choose $N$ first and then that the $\epsilon$ condition holds for any $x$. This is another way of saying that the choice of $N$ must be "independent of $x$" for uniform convergence and this is precisely why the condition fails for $f_n = \displaystyle \frac{x}{n}$.
A: $f_n$ is an odd function. Let us prove that the sequence does not converge uniformly at $[0,+\infty)$.
$$\sup_{x\in\mathbb R^+}f_n(x)\ge f_n(n)=1$$
thus
$$\lim_{n\to+\infty}\sup_{\mathbb R^+}f_n\neq 0$$
A: Geometric idea: as $f_n\to 0$ pointwise, the uniform convergence will imply that for all $\epsilon > 0$ the graph of $f_n$ is contained in the band $\Bbb R\times(-\epsilon,\epsilon)$ for $n$ large enough. But this is false for all $\epsilon$ and all $n$ because the graph of $f_n$ is a straight line with slope $1/n\ne0$.
A drawing done with Desmos (https://www.desmos.com/calculator):

