<s>[INST] Uniform convergence of $f_n(x) = \frac{nx}{1+nx^2}$ Can someone please verify my proof?

Let $f_n(x) = \displaystyle{\frac{nx}{1+nx^2}}$
(a) Find $f(x) = \lim f_n(x)$
(b) Does $f_n \longrightarrow f$ uniformly on $[0,1]$?
(c) Does $f_n \longrightarrow f$ uniformly on $[1, \infty)$?

(a) $f_n(0) = 0 $ for all $n \in \mathbb{N}$, so $f(0)=0$.
Also, if $x \neq 0$,
\begin{eqnarray}
f_n(x) &=& \frac{nx}{1+nx^2} \\
&=& \frac{1}{x+\frac{1}{nx}}
\end{eqnarray}
Using the fact that $$\lim_{n \longrightarrow \infty} \frac{1}{nx} = 0$$
we get $$\lim f_n(x) = \frac{1}{x}$$ for $x \neq 0$.
(b) Since $f$ is discontinuous at $x = 0$, it is clearly not the case that $f_n \longrightarrow f$ on $[0,1]$.
(c) Yes. Pick $n > \frac{1}{\epsilon}$. Then,
\begin{eqnarray}
n &>& \frac{1}{x^2 \epsilon}\\
nx^2 &>& \frac{1}{\epsilon} \\
1 + nx^2 &>& \frac{1}{\epsilon} \\
\frac{1}{1+nx^2} &<& \epsilon \\
\frac{1}{x(1+nx^2)} &<& \epsilon \\
\left|\frac{1}{x(1+nx^2)} \right| &<& \epsilon \\
\left|\frac{nx}{1+nx^2} - \frac{1}{x}\right| &<& \epsilon \\
|f_n(x)-f(x)| &<& \epsilon
\end{eqnarray}
Note that in arriving at some of the inequalities, the fact that $x \geq 1$ has been used.
 [/INST] c)
$sup|f_{n}(x)-f(x)|=|f_{n}(1)-f(1)|=\frac{1}{n+1}\longrightarrow 0$ as $n\longrightarrow\infty$
</s>