Limits under integrals and uniform convergence. I'm asked to discuss uniform and pointwise convergence of this sequence:
$$f_n(x)=\frac{1+nx}{2+nx^2}$$
It's easy to see that
$$\lim_{n \to \infty}f_n(x)=\frac{1}{x} \qquad \forall x \in \mathbb{R}\setminus\{0\}$$
and that $f_n(0)=\frac{1}{2}$ so we have that:
$$f_n \to f(x)=\begin{cases}\frac{1}{2} \qquad x=0 \\
\frac{1}{x} \qquad x \neq 0 \end{cases}$$
and clearly the convergence can't be uniform on $\mathbb{R}$ as $f_n(x)$ is continuous for all $n \in \mathbb{N}$ and $f(x)$ is not.
Now I try to study where it converges uniformely.  We can easily note that $f_n(x) \geq f_{n+1}(x)$ if $x\in (-\infty,0] \cup[2,+\infty) = A$ so the function converges uniformely in every compact contained A. As converse, $f_n(x) \leq f_{n+1}(x) \ \forall x \in A^C$ so we can declare that $f_n$ converges uniformely to $f$ in $\mathbb{R}\setminus\{0\}$.

  
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*First question: Until now is everything true and correct? I am not convinced about the last part.
  

Now I am asked to evaluate
$$\lim_{n\to \infty} \int_0^{\infty} \frac{1+nx}{2+nx^3}$$
as before, we have uniform convergence at $f=\frac{1}{x^2}$ in $\mathbb{R} \setminus \{0\}$ (if my previous solution was correct) and so we can divide the integral into two pieces: the one in $[a,+\infty)$ and the other one in (0,a]. As we can't talk about uniform convergence in $0$, how am I supposed to act? I intuitively think that as $f$ is $\frac{1}{x^2}$ that I'm expected to find that the limit does not exist but I'm not sure about it (the first piece, $[a,+\infty)$ is convergent, the other one I don't know).
Thanks in advance!
Thanks in advance.
 A: You are correct that $f_n(x)=\frac{1+nx}{2+nx^2}$ does not converge uniformly on $\mathbb{R}$.  But, the convergence is not uniform on $\mathbb{R}\setminus\{0\}$, as you have asserted.  
Rather, $f_n(x)$ converges uniformly on all compact sets that do not contain $\{0\}$.  So, for all $\varepsilon_1>0$ and $\varepsilon_2>0$, we have uniform convergence on compact subsets of $(-\infty,-\epsilon_1]\cup [\varepsilon_2,\infty)$.
The reason that the convergence fails to be uniform on, say $(0,2]$ is that for $x\in (0,2]$
$$\limsup_{n\to \infty}\left|\frac{1+nx}{2+nx^2}-\frac1x\right|\ne 0$$ 
Note that for $x\in [\varepsilon_2,2]$ we have
$$\limsup_{n\to \infty}\left|\frac{1+nx}{2+nx^2}-\frac1x\right|=0$$
and the convergence is uniform on $[\varepsilon_2,2]$.

Now, let's have a look at the limit  
$$\lim_{n\to\infty}\int_0^\infty \frac{1+nx}{2+nx^3}\,dx$$
The integral converges for any $n\in \mathbb{N}$, $n>0$.  Furthermore, we can write
$$\begin{align}
\int_1^\infty \frac{1+nx}{2+nx^3}\,dx&=\int_1^\infty \frac1{x^2}\,dx+\int_1^\infty \frac{x^2-2}{x^2(2+nx^3)}\,dx\\\\
&=1+\int_1^\infty \frac{x^2-2}{x^2(2+nx^3)}\,dx\\\\
\end{align}$$
It is easy to see that 
$$\lim_{n\to \infty}\left|\int_1^L \frac{x^2-2}{x^2(2+nx^3)}\,dx\right|\le \lim_{n\to \infty}\frac1n \int_1^L \frac{|x^2-2|}{x^5}\,dx=0$$
So that we assert 
$$\lim_{n\to\infty}\int_1^\infty \frac{1+nx}{2+nx^3}\,dx=1$$
However, note that
$$\int_0^1 \frac{1+nx}{2+nx^3}\,dx\ge \frac12+\frac14n$$
And so, $\lim_{n\to \infty}\int_0^1 \frac{1+nx}{2+nx^3}\,dx=\infty$
