$f_n(x) = p(x)+\frac{1}{n}$ converges uniformly to $p$ but $f_n^2$ does not converge unif. to $p^2$ 
Let $p:\mathbb{R}\to\mathbb{R}$ be a polynomial of degree $\ge 1$. Show that the sequence $f_n:\mathbb{R}\to\mathbb{R}$ given by $f_n(x) = p(x)+\frac{1}{n}$ converges uniformly to $p\in\mathbb{R}$, but $f_n^2$ does not converge uniformly to $p^2$.

In order to show that $p(x)+\frac{1}{n}$ converges uniformly to $p(x)$, I did:
$$\left|p(x)+\frac{1}{n}-p(x)\right|<\epsilon$$
just to see that $\left|\frac{1}{n}\right|<\epsilon$
which is easy to show that converges uniformly. Is this reasoning correct?
Now, for $f_n^2$, I tried $f_n^2 = p^2(x)+2p(x)\frac{1}{n}+\frac{1}{n}$ but I don't know how to manipulate it. Does somebody have an idea?
I also tried to see $p^2(x)+2p(x)\frac{1}{n}+\frac{1}{n} = p(x)(p(x)+\frac{2}{n})+\frac{1}{n}$ and I can at least see that it'll converge pointwise to $p^2(x)$, but how to prove it's not uniformly?
 A: Let $ p(x)=a_0+a_1x+a_2x^2+\ldots +a_nx^n;\deg p(x)\ge1\implies \text{ at least one }a_i \text{for} i\ge 1 \text {is}\neq 0$
Now $(f_n(x))^2=(p(x))^2+\frac{1}{n^2}+2\dfrac{p(x)}{n}$
Now $|(f_n(x))^2-(p(x))^2|=2\dfrac{p(x)}{n}+\dfrac{1}{n^2} =\dfrac{2(a_0+a_1x+a_2x^2+\ldots +a_nx^n)}{n}+\dfrac{1}{n^2} $ .
For $x=n$ we have $|(f_n(x))^2-(p(x))^2|$
$=\dfrac{2a_0}{n}+2(a_1+a_2n^2+\ldots+a_nn^n)+\dfrac{1}{n^2} $
\begin{cases} \to 2a_1 & \text{if $\deg p(x)=1$}\\ \to \infty & \text{if $\deg p(x)>1$}\end{cases} 
Hence convergence is not uniform.
A: Observe that
$$
\left|\left(p(x)+\frac{1}{n}\right)^2-p(x)^2\right|=\left|\frac{2p(x)}{n}+\frac{1}{n^2}\right|=\frac{1}{n}\left|2p(x)+1/n\right|\geq\frac{1}{n}\left|2|p(x)|-\frac{1}{n}\right|,
$$
where I've used the reverse triangle inequality in the last step. If $p$ is not a constant then it must satisfy $|p(x)|\to+\infty$ as $|x|\to+\infty$, but then we can't have uniform convergence as this implies
$$
\sup_x\left|\left(p(x)+\frac{1}{n}\right)^2-p(x)^2\right|\geq\sup_x\frac{1}{n}\left|2|p(x)|-\frac{1}{n}\right|=+\infty.
$$
A: You can use Cauchy's criterion. Just observe that $\left| f_n(x)^2-f_m(x)^2\right|=\left|\frac{(m-n)(m+n+2mnp(x))}{m^2n^2}\right|$ and you can take $x$ large enough so that this is greater than $\epsilon$ whatever $n$ and $m$ are.
