Showing pointwise and uniform convergence of $\frac{nx^2}{1+n^2x}$ Suppose we have:

$$f_n: \mathbb{R}^{+} \rightarrow \mathbb{R} $$ $$ f_n(x)=\frac{nx^2}{1+n^2x}$$

I started of by showing the pointwise convergence.
For $x=1$, we have: 
$$f_n(1)=\frac{n}{1+n^2} \leq \frac{1}{n}$$ thus if $f=0$, then $f_n$ converges to $f$.
For $x < 1$ we have:
$$x^2 < x \implies nx^2 < n^2x \implies lim_{n \rightarrow \infty} = 0$$
Thus if $f=0$, then $f_n$ converges to $f$.
For $x>1$, we have:
$$f_n(x)=\frac{nx^2}{1+n^2x}=\frac{n}{\frac{1}{x^2}+\frac{n^2}{x}} \leq \frac{n}{ \frac{n^2}{x} }=\frac{nx}{n^2}=\frac{x}{n}$$, thus if $f=0$, then $f_n$ converges to $f$.
In all cases $f_n$ converges to $0$. So it pointwise converges to $0$.
Now I want to study the uniform convergence of the function. And this is where I am stuck.  $f_n(x) \leq \frac{1}{n}$ if $x \leq 1$, and otherwise we have $f_n(x) \leq \frac{x}{n}$.
So if I put $a_n= \frac{x+1}{n}$, we have then $||f_n -f||_{\infty} \leq a_n$, and $lim_{n \rightarrow \infty}=0$, thus $f_n$ convergence uniformely. Am I correct?

uniform convergence: you can rewrite $f_n$ as follows :
$$ \begin{array}[rcl] ~f_n(x) &=& \frac{nx^2}{1+n^2x} \\
&=& \frac{x^2}{\frac{1}{n^2}+x}\frac{1}{n} \end{array}$$ 
If $\bar{x} = n^2 \in \mathbb{R}^+$, we have
$$ f_n(\bar{x}) = \frac{n^4}{\frac{1}{n^2}+n^2}\frac{1}{n} = \frac{n^3}{\frac{1}{n^2}+n^2} \rightarrow  +\infty $$
Hence, the sequence cannot converge uniformly for $x \in \mathbb{R}^+$.  
