Pointwise but not Uniformly Convergent The Question: Prove that the sequence of functions $f_n(x)=\frac{x^2+nx}{n}$ converges pointwise on $\mathbb{R}$, but does not converge uniformly on $\mathbb{R}$. 
My Work: Prove Pointwise: First, $\lim\limits_{n\to\infty} \frac{x^2+nx}{n}=\lim\limits_{n\to\infty} \frac{x^2}{n}+x=x$. 
My Problem: I am not sure where this fails to be uniformly convergent. Any help is appreciated. 
Thanks
 A: What is $\sup\{|f_n(x)-f(x)|\}\to?$ as $n\to\infty$?
A: Looking at the form of $f_n(x) = {x^2+nx\over n} = {x^2\over n}+x$, it is reasonable to propose $f(x)=x$ as the pointwise limit. To verify this, note that $f_n(x)\to f(x)$ pointwise on an interval $I$ means
$$
|f_n(x)-f(x)|\to 0\text{ as }n\to \infty \text{ for each fixed }x\in I.
$$
With that in mind, consider 
$$
|f_n(x)-f(x)|=\left|\left({x^2\over n}+x\right)-x\right|=|x^2/n|\to 0\text{ as }n\to\infty\text{ for each fixed }x\in\mathbb R.
$$
We conclude that indeed $f_n(x)\to x$ pointwise on $\mathbb R$.
On the other hand, $f_n(x)\to f(x)$ uniformly on $I$ means
$$
\sup_{x\in I}|f_n(x)-f(x)|\to 0\text{ as }n\to \infty.
$$
Examining this (stronger!) condition in our case, we see
$$
\sup_{x\in\mathbb R}|f_n(x)-f(x)|=\sup_{x\in\mathbb R}\left|\left({x^2\over n}+x\right)-x\right|=\sup_{x\in \mathbb R}|x^2/n|\geq |n^2/n|=n\not\to 0\text{ as }n\to\infty.
$$
Thus, $f_n(x)$ does not not converge uniformly to $f(x)$ on $\mathbb R$.
A: Choose $\epsilon > 0$. If the convergence is uniform, then you can find an $n$ such that $|\frac{x^2}{n} + x -x|_{\infty} = |\frac{x^2}{n}|_{\infty}$ is smaller than $\epsilon$. That is, there exists an $n$ such that for ALL $x$, $x^2/n$ is smaller than $\epsilon$. Is this possible?
A: I remember that converging sequences are always bounded, that is, there is a metric space structure on which convergence is defined. In this time the elements of the given sequence don't have distances. Hence there cannot be convergence.
A: A sequence of functions $f_{n}:\mathbb{R}\rightarrow\mathbb{R}$ is said to non-uniformly contionous on $\mathbb{R}$ , if $\exists \epsilon_{0}>0:\exists$ subsequences  $n_{k},x_{k}$,such that 
$|f_{n_{k}}(x_{k})-f(x_{k})|\ge\epsilon_{o}   \forall  k\in \mathbb{N}$
From a simple calculation,we have that $f_{n}(x)\rightarrow x,\forall x\in\mathbb{R}$
Now,define $n_{k}:=k,x_{k}:=\sqrt{k}, \forall k\in \mathbb{N}$,so that,
$|f_{n_{k}}(x_{k})-f(x_{k})|=|\frac{k}{k}+k-k|=1>\epsilon_{0},$ for a certain $\epsilon_{0}<1$
