Show that $(f_n)^2$ doesn't converge uniformly to $p^2$". Please,check my solution to item "a" and give me a hint to solve item "b". Thanks in advance.
Problem:
"Let $p:\mathbb {R} \rightarrow \mathbb R$ be a polynomial whose degree is $\ge 1$. (a) Show that the sequence of functions $f_n:\mathbb {R} \rightarrow \mathbb R$,given by $f_n(x)= p(x) + \frac{1}{n}$ uniformly converges to $p(x)$ in $\mathbb R$. (b) Show that $(f_n)^2$ doesn't converge uniformly to $p^2$".
solution item "a": I must prove that given $\varepsilon >0$, there exists $n_0 \in \mathbb N$ such that $n>n_0 \Rightarrow \left|p+\frac{1}{n}-p \right| < \varepsilon$, for every $x \in \mathbb R$.
$\left|p+ \frac{1}{n} -p \right |=$$\left | \frac{1}{n} \right |< \varepsilon  \Rightarrow \frac{1}{n} < \varepsilon$. So, if I choose $n_0> \frac{1}{\varepsilon}$, then $n>n_0 \Rightarrow  \left| f_n - p \right |< \varepsilon$. It proves that $f_n$ uniformly converges to $p$.
Solution item "b": I must prove that there exists $\varepsilon >0$ shuch that, for every $n \in \mathbb N$, $\left | p^2+\frac{2p}{n}+\frac{1}{n^2}-p^2 \right |\ge  \varepsilon$. How should I proceed to find $\varepsilon$?
 A: Nope, you have to proof that there exists $\epsilon>0$ such that for any $n$ there exists $x_n\in \mathbb{R}$ such that 
$$\left| p(x_n)^2+\frac{2p(x_n)}{n}+\frac{1}{n^2}-p(x_n)^2\right|>\epsilon$$
The expression in the absolute value is
$$\frac{2p(x)}{n}+\frac{1}{n}$$
and you know that $\deg p(x)\geq 1$, meaning that 
$$\lim_{x\to\pm\infty} |p(x)|=+\infty\;.$$
You should use this to show that the expression under consideration has to be bigger than some fixed $\epsilon$ at some point $x_n$, no matter what $n$ you pick.
A: Your solution to item $a)$ is correct. 
To solve $b)$ you have: $\left | p^2(x)+\frac{2p(x)}{n}+\frac{1}{n^2}-p^2(x) \right|=\left|\frac{2p(x)}{n}+\frac{1}{n^2}\right|=\frac{1}{n}\left|2p(x)+\frac{1}{n}\right|, \forall x\in\mathbb{R}.$ Since the degree of $p$ is $\ge 1,$ for any $n\in\mathbb{N},$ there exists $x_n$ such that $|p(x_n)|\ge n^2.$ So,
$$\left | p^2(x_n)+\frac{2p(x_n)}{n}+\frac{1}{n^2}-p^2(x_n) \right|=\frac{1}{n}\left|2p(x_n)+\frac{1}{n}\right|\ge 2n^2-\frac1n\ge 1.$$ This shows that the convergence is not uniform.
A: It looks like you have the basic idea of part (a) down just fine, although I have a couple remarks. When you state things like "$n>n_0 \Rightarrow  \left| f_n - p \right |< \varepsilon$" I would instead use $n>n_0 \Rightarrow  \left| f_n(x) - p(x) \right |< \varepsilon$ because the way you stated it refers to the functions $f,p$ when we are really just interested in the values $f(x), p(x).$ This is extra important for proofs about uniform continuity, because these inequalities have to hold for all $x$ in the domain of $f$ and $p$. Second, when you get to the line $\left|\frac{1}{n}\right|< \varepsilon$, it feels like you have already assumed that the function converges uniformly. I would state the proof more like: 
"Let $\varepsilon>0$. Then we know there exists $n_0 \in \Bbb{N}$ such that $\varepsilon>\left|\frac{1}{n_0}\right|$. Further, for any $n>n_0$ we know $$\varepsilon>\left|\frac{1}{n_0}\right|>\left|\frac{1}{n}\right|=\left|\frac{1}{n}+0\right| = \left|\frac{1}{n}+(p(x)-p(x))\right|= \left|\left(\frac{1}{n}+p(x)\right)-p(x)\right|=\left|f_n(x)-p(x)\right|$$ Now you are guaranteed uniform convergence. 
As for a hint on part (b), it looks like many other users have offered hints and I have nothing to say that hasn't been said yet. Good luck!
