Issues Identifying Uniform Norm of a Sequence of Functions? 
3. Let $f_{n}(x):=\frac{\cos \left(n^{2} x\right)+2 n x^{3}}{(n+5) x^{4}+1}$ for $x \in \mathbb{R}$ and $n \in \mathbb{N}$
(a) Find $f(x):=\lim _{n \rightarrow \infty} f_{n}(x)$ for all $x \in \mathbb{R}$
(b) Find $\lim _{n \rightarrow \infty} \int_{1}^{2} f_{n}(x) d x .$ Fully justify your claim.
(c) Does $\left\{f_{n}\right\}$ converge uniformly to $f$ on $[-1,1] ?$ Justify your claim.

(OCR'd from picture: Problem 3)
I'm trying to solve part b of this problem, and in order to interchange the limit and integral, the sequence of functions has to be uniformly convergent on the interval [1,2]. I've gotten that $\{f_n(x)\}$ converges pointwise to $f(x)=\frac{2}{x}$ over [1,2], but cannot prove that $f_n(x)$ converges uniformly on the interval. I attempted to use the definition of the uniform norm, with $$\Vert{f_n(x)-f(x)}\Vert := \sup_{x\in\Bbb{R}}\vert f_n(x)-f(x)\vert$$
and that isn't helping... Any help on the matter would be greatly appreciated!
 A: You don't need uniform convergence. You can use DCT. Note that $0 \leq f_n (x) \leq \frac {1+16n} {n+6} \leq {16}$. Hence the integral converges to $\int_1^{2}\frac 2 x dx=2 \ln 2$.
A: The convergence is in fact uniform. Splitting the sum above,
\begin{align}
f_n(x)=\frac{\cos \left(n^{2} x\right)+2 n x^{3}}{(n+5) x^{4}+1} 
&=\frac{\cos \left(n^{2} x\right)}{(n+5) x^{4}+1} +\frac{ 2n x^{3}}{(n+5) x^{4}+1}\\
\end{align}
The first term is uniformly decaying to $0$ for $x\in[1,2]$,
$$\left|\frac{\cos \left(n^{2} x\right)}{(n+5) x^{4}+1}\right|\le\frac{1}{(n+5)+1} \to 0.$$
For the second term, we have
$$\frac{ 2n x^{3}}{(n+5) x^{4}+1}-\frac2x=\frac{ 2 x^{3}}{ x^{4} + \frac{5x^4}n+\frac1n} - \frac {2x^3}{x^4}  $$
Note that these terms abstractly look like $2x^3$ times a finite difference,  $$\left|\frac{1}{y+h}-\frac1{y}\right| = \frac{h}{y(y+h)} \le \frac{h}{y^2}, $$
where $h=\frac{5x^4}n+\frac1n>0$ is small. Applying this,
$$\left|\frac{ 2n x^{3}}{(n+5) x^{4}+1}-\frac2x\right| \le 2x^3\frac{\frac{5x^4}n+\frac1n}{x^8} = \frac{1}{n}\cdot\frac{2(5x^4+1)}{x^5} \le\frac{ 2(5\cdot 2^4+1)}n \to 0$$
Since all bounds are uniform in $x$, the triangle inequality gives
$$\sup_{x\in[1,2]} \left|f_n(x) - \frac2x\right| \to 0.$$
