Uniform convergence of $g_n(x) := \frac{(x+1)^2 e^{-x} \sin x}{n^2x^2+n\sqrt n x +1}$ over $[0, +\infty)$

Question

This is part of a larger exercise: does $$g_n : [0, +\infty) \to \mathbb R ,\ g_n(x) := \frac{(x+1)^2 e^{-x} \sin x}{n^2x^2+n\sqrt n x +1}$$ converge uniformly for $x\ge0$?

The sequence of functions converges pointwise to $0$, which is quite immediate. For uniform convergence, I tried elementary inequalities, but nothing useful when you have to write something like $\sup_{x \ge 0} \lvert f_n(x) \rvert \le \cdots$ Deriving does not seem useful either. How would you answer?

Answer 1

For uniform convergece on $[0,1]$ use the inequality $n^{2}x^{2}+n\sqrt n x+1\ge n\sqrt n x$ and the fact that $\frac {\sin x } x$ is bounded.

For uniform convergece on $[1,\infty]$ use the inequality $n^{2}x^{2}+n\sqrt n x+1\ge n^{2}x^{2}$.