I'm trying to show that $f_n(x) = (x + \frac{1}{n})^2$ converges uniformly. So $f(x) = \lim_{n \to \infty} f_n(x) = \lim_{n \to \infty} (x + \frac{1}{n})^2 = (\lim_{n \to \infty} (x + \frac{1}{n}))^2 = x^2$.
I want to show that $|f_n(x) - f(x)| < \epsilon$ for $n$ large enough. So
$|f_n(x) - f(x)| = |(x + \frac{1}{n})^2 - x^2| = |x^2 + \frac{2x}{n} + \frac{1}{n^2} - x^2| = |\frac{2xn}{n^2} + \frac{1}{n^2}|$
Edit Uniform convergence does not occur on $\mathbb{R}$. But let me consider the interval $[-5, 5]$. Since $x \in [-5, 5]$, $|\frac{2xn}{n^2} + \frac{1}{n^2}| < |\frac{10n}{n^2} + \frac{1}{n^2}|$, which will be made less than $\epsilon$ for $n$ large enough.