I know this sequence of functions converges to $0$ pointwise, so I have to show that for all $\epsilon > 0$, there exists an $N$ such that for all $n > N$, $d(\frac{nx^3}{1 + n^2 x^2}) < \epsilon$. I can see that $d(\frac{nx^3}{1 + n^2 x^2}) < d(\frac{nx^3}{n^2 x^2}) = d(\frac{x}{n})$, so I can't take this approach since $\frac{x}{n}$ is not uniformly convergent (I think I have to show something like $d(\frac{nx^3}{1 + n^2 x^2}) < \epsilon < d(\frac{nx^3}{n^2 x^2}) = d(\frac{x}{n})$.)
I thought it might not converge uniformly so I tried taking the derivative and setting it equal to $0$, but the result is that $x = 0$ so that doesn't work since the sequence does converge pointwise to $0$.
What other approaches can I take?
The sequence does not converge uniformly to $0$ on $[1,\infty)$. Consider what happens when $x=n$.