Uniform Convergence of $f_n(x) = \frac{n(1-2x) + \cos x -x\sin^2 x}{2n + \sin^2 x}$ [proof verification] As the title suggests, I'm trying to show that $f_n(x) = \frac{n(1-2x) + \cos x -x\sin^2 x}{2n + \sin^2 x}$ converges uniformly. I tried showing that $f_n(x)$ converges uniformly to the pointwise limit $\frac{1-2x}{2}$ but I couldn't get it to work out, so I've tried to show it's uniformly Cauchy like so:
Let $f_n(x) = \frac{n(1-2x) + \cos x -x\sin^2 x}{2n + \sin^2 x}$.
Let $m, n \in \mathbb{N}$ such that $m > n$. Consider
$$|f_n(x) - f_m(x)| = |\frac{(m-n)(2\cos(x) - \sin^2(x))}{(2m+\sin^2(x))(2m+\sin^2(x))}| \leq \frac{|m-n|(2)}{4mn} = \frac{|n - m|}{2mn} < \frac{n}{2n^2} < \frac{1}{2n} < \epsilon$$
So let $\epsilon > 0$.
Choose $N \in \mathbb{N}$ such that $N > \frac{1}{2\epsilon}$.
Then for all $m > n > N$,
$$|f_n(x) - f_m(x)| < \epsilon$$
So $f_n$ is uniformly Cauchy, so it must converge uniformly.
But after checking this result graphically I think that it's wrong, and I'm not really sure what else to do.
Any help would be greatly appreciated.
 A: The convergence is uniform for all $x \in \mathbb{R}$.
The pointwise limit is clearly $f(x) =(1-2x)/2$ and proceeding directly,
$$|f_n(x) - f(x) | = \left|\frac{\frac{1-2x}{2} + \frac{\cos x}{2n}- \frac{x\sin^2x}{2n}}{1 + \frac{\sin^2x}{2n}}- \frac{1-2x}{2} \right| \\ = \frac{\left|\frac{1-2x}{2} + \frac{\cos x}{2n}- \frac{x\sin^2x}{2n}- \frac{1-2x}{2}- \frac{\sin^2x}{2n}\frac{1}{2} + \frac{\sin^2 x}{2n} \frac{2x}{2}\right|}{1 + \frac{\sin^2x}{2n}} \\ \leqslant \left|\frac{\cos x}{2n}- \frac{\sin^2x}{4n} \right|\leqslant \frac{3}{4n}\underset{n \to \infty}\longrightarrow 0$$

Checking your approach, with $m >n$, we have
$$f_m(x) - f_n(x) \\=\frac{m(1-2x)2n + m(1-2x)\sin^2x + (\cos x - x\sin^2x)2n + (\cos x - x\sin^2x)\sin^2 x}{(2m+\sin^2x)(2n+ \sin^2 x)} \\ -\frac{n(1-2x)2m + n(1-2x)\sin^2x + (\cos x - x\sin^2x)2m + (\cos x - x\sin^2x)\sin^2 x}{(2m+\sin^2x)(2n+ \sin^2 x)} \\ =  \frac{(m-n)(1-2x)\sin^2 x +2(n-m)(\cos x - x \sin^2 x)}{(2m+\sin^2x)(2n+ \sin^2 x)}\\ =  \frac{(m-n)(\sin^2 x -2\cos x)}{(2m+\sin^2x)(2n+ \sin^2 x)},$$
and, thus,
$$|f_m(x) - f_n(x)| = \frac{(m-n)|\sin^2 x -2\cos x|}{(2m+\sin^2x)(2n+ \sin^2 x)}\leqslant \frac{3(m-n)}{4mn} = \frac{3}{4n} - \frac{3}{4m} < \frac{3}{4n}$$
For all $m > n > N(\epsilon)$ where $N(\epsilon) > \frac{4}{3\epsilon}$ we have $|f_m(x) - f_n(x)| < \epsilon$.
