I have found that the limit function of pointwise convergence $f(x)$ is equal to $f(x) = \frac{2x}{1+x^4}$. I am having difficulty showing that $\{f_n\}$ converges uniformly to $f$ on $\mathbb{R}$. I worry that I am missing something conceptually about convergence, but the logic I am intending to use is as follows:
Take $\lvert f_n - f \rvert = \Bigl\lvert \frac{2nx}{n + nx^2 + \cos(nx)} - \frac{2x}{1+x^4} \Bigr\rvert$. Then take $\lim\limits_{n \rightarrow \infty} (\sup\limits_{x \in \mathbb{R}} \lvert f_n - f \rvert )$ to prove uniform convergence. Taking the limit here is gross and makes me think that I am supposed to find an easier function bounding $\lvert f_n - f \rvert$ above. But I am having a hard time finding any such function; anything that I can come up with doesn't simplify the process. I may be missing something obvious, but I'm not sure. In short, I am a little stuck as to how to proceed with proving the uniform convergence of $\{f_n\}$. What am I missing?
Note that $$f_n(x)-f(x)=\frac{2x}{1 + x^4 + \frac{\cos(nx)}{n}} - \frac{2x}{1+x^4}= \frac{-\frac{2x\cos(nx)}{n}}{(1 + x^4 + \frac{\cos(nx)}{n})(1+x^4)}.$$ Hence for $n\geq 2$, $|\frac{\cos(nx)}{n}|\leq \frac{1}{2}$ and $$|f_n(x)-f(x)|\leq\frac{\frac{2|x|}{n}}{(1 + x^4 - \frac{1}{2})(1+x^4)}=\frac{2M}{n}$$ where $M=\max_{x\in\mathbb{R}}\frac{|x|}{(\frac{1}{2} + x^4 )(1+x^4)}$ (it is a bounded function!). It follows that $$\lim\limits_{n \rightarrow \infty} (\sup\limits_{x \in \mathbb{R}} \lvert f_n - f \rvert )\leq \frac{2M}{n}\to 0.$$