• For $f_n(x)=\sin^n(x)$ I think your solution is good.
• For $f_n(x)=\frac{x^n}{n}+1$, if we define $d_n$ as above, note that $0\leq d_n\leq 1/n$ (because $0\leq x<1$ or simpler: because the application $x\mapsto x^n/n$ is an increasing function of $x$). This implies that $d_n\to 0$
• For $f_n(x)=\frac{1}{1+(x-n)^2}, x<0$, note that the restriction over $x$ implies that $n^2<1+(n-x)^2$ and therefore, that $f_n(x)<n^{-2}$. Using the same argument used in the last point, we have that the convergence is uniform.
• For $f_n(x)=\frac{1}{1+(x-n)^2}, x>0$ it's fine.
• For $f_n(x)= nxe^{-nx}$ it's fine.
• For $f_n(x)=x^n, x\in [0,1]$ it's fine.
• For $f_n(x)= \frac{\sin(nx)}{\sqrt{n}}$ on $\mathbb{R}$ it's fine.
• For $f_n(x)=\frac{nx}{1+nx}, x\in (0,1)$ the convergence is not uniform since for every $N$ there exists an $n\geq N$ and an $0<x<1$ such that $\vert f_n(x)-f(x)\vert \geq 1/4$ : just take $x=1/n$ and note that $\vert f_n(1/n)-f(1/n)\vert=1/2$.
• For $f_n(x)=\frac{x^n}{1+ x^n}, x\in [0,2]$ it's fine.
• For $f_n(x)=n^2x^2e^{-nx}$ over $(0,\infty)$ it's fine.
• For $f_n(x)=(\cos(\pi n!x))^{2n}, x\in [0, 1]$ your conclusion is correct, but the limit function is not the one that you said: note that if $x$ is a rational number, for $n$ large enough we will have $n!x\in\mathbb{Z}$ and then, $(\cos(\pi n!x))^{2n}=1$. Then, the limit is
$$
f(x)=\begin{cases}
1 & \mathrm{if} \quad x \quad \mathrm{rational}\\
0 & \mathrm{if} \quad x \quad \mathrm{not \, rational}
\end{cases}
$$
Since $f$ is clearly not continuous, the convergence is not uniform.
• For $f_n(x)=n^2x(1-x^2)^n$ I think it is fine too
I hope this will be useful for you
I don't believe the statement you're trying to prove is correct; the result does not hold on the unbounded interval $(0, \infty)$. If this interval was bounded, this result would hold, as you have already noted.
For a concrete example, consider $f(x) = x + 1$ and the sequence of functions $\{f_n\}$ defined by $$f_n(x) = x + \frac{n - 1}{n}.$$ Then clearly, $\{f_n\}$ converges uniformly to $f$. On the other hand, we have $$\lvert F(x) - F_n(x) \rvert = \bigg\lvert \int_0^x f(t) \, dt - \int_0^x f_n(t) \, dt \bigg\rvert = \bigg\lvert \int_0^x \frac 1n \, dt \bigg\rvert = \frac xn$$ for all $n$, which is larger than any $\varepsilon > 0$ for sufficiently large $x$.
Best Answer
The convergence is not uniform. For example consider the sequence $x_n=n\pi$. Then, \begin{align} \bigl|f_n(x_n)-f(x_n)|&=\bigl|\cos(n+1)\pi-\cos n\pi\bigr|\\ &=\bigl|\cos n \pi\cos \pi-\cos n\pi\bigr|\\ &=2|\cos n \pi|=2 \end{align} If the convergence was uniform, then for $\epsilon=1$ we would be able to find $N$ such that $$\sup_{x\in \mathbb{R}}\bigl|f_n(x)-f(x)|\leq 1$$ for every $n\geq N$. But this contradicts the fact that $|f_n(x_n)-f(x_n)|=2>1$ for all $n$.