Uniform convergence on two separate intervals

Test the sequence of functions for uniform convergence

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

$a) \text{on} \ [0,1]$

$b) \text{on} \ [1,{\infty})$

a)

If we differentiate this with respect to x we get the following :

$$(\frac{nx}{1+n^2x^2})'=\frac{n(1-n^2x^2)}{(1+n^2x^2)^2}$$

This is equal to $0$ for $x=^{+}_{-}\frac{1}{n}$

From here we can either check these two values in the original function and take the larger one as the supremum or we can find the second derivative and check whether it's negative. In either case we find that for $x=\frac{1}{n}$ we get the value $f_n(x)=\frac{1}{2}$ and for $x=-\frac{1}{n}$ we get $f_n(x)=-\frac{1}{2}$. Note that $x=-\frac{1}{n}$ doesn't even belong in our interval. Now we can finish the problem.

$$\lim_{n\to \infty}{\frac{nx}{1+n^2x^2}}=0$$

So it is piecewise converging to $f(x)=0$

$$\lim_{n\to \infty}\sup{\left \lvert{f_n(x)-f(x)}\right \rvert}=\lim_{n\to \infty}{\frac{n\frac{1}{n}}{1+n^2 \frac{1}{n^2}}}=\frac{1}{2}$$

So $f_n(x)$ does not convergence uniformly on $[0,1]$ to $f(x)=0$

b)

Now we need to test the on the interval $[1,\infty)$.From the first derivative we notice that if $x\in [1,\infty)$ then the numerator seems to be negative for all $x>1$. So $f_n(x)$ is monotonically decreasing. That means that the supremum will be at the first point of the interval or $x=1$. Taking the limit :

$$\lim_{n\to \infty}{\frac{n}{1+n^2}}=0$$

We can see that the function convergences uniformly on this interval.

Is there something wrong with either of these solutions?

I'm having doubts mostly about the second interval since $x=\frac{1}{n}$ is a supremum but only for $n=1$. However I believe that we need to have the supremum that's satisfied for all $n \in N$. Pointing even the tiniest mistake in my solutions would be very helpful. Thank you.

You sequence does not converge uniformly on $[0,1]$. It converges pointwise to the null function but, since $(\forall n\in\mathbb{N}):f_n\left(\frac1n\right)=\frac12$, the convergence is not uniform.

However, it converges uniformly to the null function on $[1,+\infty)$. Your proof of this fact is correct.