$f_n\cdot g_n$ not always uniformly converges I've been asked to prove that for $f_{n}\left(x\right)=g_{n}\left(x\right)=x-\frac{1}{n}$ in domain $\mathbb{R}$ that $f_n\cdot g_n$ isn't uniformly converges. 
So first I've showed that $f$ and $g$ pointwise converges to $x$ easly.
But i cant understand how to prove the rest, beside starting from $\left|f_{n}\cdot g_{n}\left(x_{n}\right)-f\cdot g\left(x_{n}\right)\right|$. 
It would be really good if you could advise me on what I need to look at when Im defining $x_n$, $n_k$ and $\varepsilon_0$ generaly when Im facing this kind of counter uniform converges questions. 
Ok so i think im i got it, if $x_n=n$ and $\varepsilon_0=2$
$\left|f_{n}\cdot g_{n}\left(x_{n}\right)-f\cdot g\left(x_{n}\right)\right|=\left|f_{n}\left(x_{n}\right)\cdot g_{n}\left(x_{n}\right)-f\left(x_{n}\right)\cdot g\left(x_{n}\right)\right|=\left|\left(x_{n}-\frac{1}{x_{n}}\right)^{2}-x_{n}^{2}\right|=$
$=\left|x_{n}^{2}-\frac{2x_{n}}{n}+\frac{1}{n^{2}}-x_{n}^{2}\right|=\left|\frac{2n}{n}+\frac{1}{n^{2}}\right|=\left|2+\frac{1}{n}\right|>2=\varepsilon_{0}$
Is this legal proof ?
 A: Here, as you've indicated in your post, the pointwise limit is $x^2$. Recall a sequence of functions converges uniformly, the limit is the same as the pointwise limit. By definition, the sequence $f_n(x) \cdot g_n(x)$ converges uniformly to $x^2$ (on $\mathbb R$) if for each $\epsilon > 0$, there exists some $N(\epsilon)$ such that $|f_n(x) \cdot g_n(x) - x^2| < \epsilon$ for all $n \ge N(\epsilon)$ and for all $x \in \mathbb R.$ 
We consider the question: for each $\epsilon$, does there exist an $N(\epsilon)$ such that for all $n \ge N(\epsilon)$ and for all $x \in \mathbb R$, we have $|f_n(x) \cdot g_n(x) - x^2| = |\frac {2x} {n} + \frac {1} {n^2}| < \epsilon$? The answer, as you correctly identified, is no, because for any fixed $n$, there exists an $x \ge n$, so that $|f_n(x) \cdot g_n(x) - x^2| = |2 + \frac {1} {n^2}| > 2.$
MotylaNogaTomkaMazura's proof is essentially a similar idea. Note that $f_n(x) \cdot g_n(x)$ converging to $x^2$ uniformly means that $\lim_{n\rightarrow\infty} \sup_x |f_n(x) \cdot g_n(x) - x^2| = 0$.
A: $$\sup_x |(x-n^{-1} )^2 -x^2 | =\sup_x |2xn^{-1} -n^{-2} |=\infty$$
