I showed that $f_n = \dfrac{nx^2 + 4n^2x -1}{x^2 + 4nx +4n^2}$, on a compact interval $I \subset \mathbb{R}$.
The question in my mind is: what does the compact interval change? When we consider the whole real line, this is not convergent uniformly, but this is on a compact interval. What is the reason for that?
We want to look at $$ \sup_{x\in\mathbb{R}} \lvert x-\frac{nx^2+4n^2x-1}{x^2+4nx+4n^2}\rvert $$ bring everything to a common denominator $$ \sup_{x\in\mathbb{R}} \lvert \frac{{x^3+4nx^2+4xn^2}}{{x^2+4nx+4n^2}}-\frac{nx^2+4n^2x-1}{x^2+4nx+4n^2}\rvert = \sup_{x\in\mathbb{R}} \lvert \frac{x^3+2nx^2+1}{x^2+4nx+4n^2}\rvert $$ Note that for any fixed value of $n$ that the numerator is ${O}(x^3)$ but the denominator is $O(x^2)$ which means this error can grow arbitrarily large. As such the $\sup$ does not exist over all real numbers for each $n$ so over $\mathbb{R}$ this does not uniformly converge.