Investigate whether $f_n (x)=\dfrac{nx}{1+n^3x^3}$ on $[0,\infty),n\in \mathbb N$ converges uniformly or not.

Investigate whether $f_n (x)=\dfrac{nx}{1+n^3x^3}$ on $[0,\infty),n\in \mathbb N$ converges uniformly or not.

Obviously, $f_n(x)$ pointwise converges to $0$. Since $f_n(0)=0$ when $x=0$ and $\displaystyle\lim_ {n\to \infty} f_n(x)=\dfrac{nx}{1+n^3x^3}=0$ for $x \in (0,\infty)$. Therefore $f_n(x)$ can converge uniformly to $0$. To show that it indeed converges to $0$, we need to check the following.

$M_n=\displaystyle\sup_{x\in [0,\infty)} |f_n(x)-f(x)|\to 0$. It turns out that $M_n=\displaystyle\sup_{x\in [0,\infty)} |\dfrac{nx}{1+n^3x^3}-0|=\sup|\dfrac{nx}{1+n^3x^3}|$.I want to play with inequalities so as to do I checked first derivative of $\dfrac{nx}{1+n^3x^3}$ which gives when function takes its maximum but I couldn't obtain something useful. After that I tried the following simply however I couldn't get rid of the term $x$.

$M_n=\displaystyle\sup_{x\in [0,\infty)} |\dfrac{nx}{1+n^3x^3}-0|=\sup|\dfrac{nx}{1+n^3x^3}|\le\sup|\dfrac{nx}{n^3x^3}|=\sup|\dfrac{1}{n^2x^2}|$

Thanks in advance.