Pointwise but not Uniformly Convergent

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 $\mathbb R$.

What is $\sup\{|f_n(x)-f(x)|\}\to?$ as $n\to\infty$?