How to see if the convergence is uniform?

Question

I have the following : $f_n(x)=\frac{nx}{1+nx^3} \quad n=1,2,\dots \quad$ and $f(x)=\lim_{n \to \infty} f_n(x) ,$

and I have done the following : $$|f_n(x) -f(x)| = \biggl|\frac{nx}{1+nx^3}-\frac{1}{x^2}\biggr| = \biggl|\frac{1}{x^2(1+nx^3)}\biggr |$$ where $\frac{1}{x^2}$ is the convergence of the series of the function. Now I know that $\Bigl|\frac{1}{x^2(1+nx^3)}\Bigr|< \epsilon$ but now I don't know how to choose $N$ in order to fully prove that it is uniformly convergent or not. Can somebody help me proceed with the proof?

Answer 1

Hint: The $f_n$'s are bounded on $(0,\infty).$ If the convergence were uniform there, then $f$ would be bounded there.

Answer 2

Assume that $f_n \to f$ uniformly on $\langle 0,1\rangle$. For $\varepsilon = 1$ there exists $n_0 \in \Bbb{N}$ such that $$n \ge n_0 \implies |f(x)-f_n(x)| < 1.$$

Now take $n := n_0 + 1$ and consider $$\left|f\left(\frac1n\right) - f_n\left(\frac1n\right)\right| = n^2 - \frac1{1+\frac1{n^2}} \ge n^2-1 \ge n_0^2 \ge 1$$ which is a contradiction. This implies that the convergence is not uniform on $\langle 0,1\rangle$.