Prove that for $f: [0, 1] \to \mathbb{R} , f_n(x) = \frac{nx}{1+n^2 x^2}$ is uniform convergent.
I found the pointwise limit to be 0. Taking the first derivative, $$f_n '(x) = \frac{n(1-n^2 x^2)}{(1+n^2 x^2)^2} $$ $f_n '(x) = 0$ at $x = \frac{1}{n}, 0, $ and $ \frac{-1}{n}$. $f_n(\frac{1}{n}) = \frac{1}{2}, f_n(\frac{-1}{n}) = \frac{-1}{2}, $ and $ f_n(0) = 0 $. Thus $f_n(\frac{1}{n}) = \frac{1}{2}$ is the max of the function.
Thus, $sup_{x \in [0,1]} |f_n(x) - 0| < 1/2 $
I do not know how to proceed in order to prove uniform convergence.
Clearly, the convergence is not uniform on $[0,1]$ since $1/n \in [0,1]$ and as $n \to \infty,$
$$\sup_{x \in [0,1]} |f_n(x)-0| = \sup_{x \in [0,1]} \frac{nx}{1 + n^2x^2} \geqslant \frac{n \cdot \frac{1}{n}}{1 + n^2 \cdot \frac{1}{n^2}} = \frac{1}{2} \not\to 0$$