Originally Posted by

**Moo** So we were left to :

$\displaystyle |f_n(x)-f(x)|=\frac{1}{x(1+nx^2)}$

For $\displaystyle x \in (0, \infty)$, there exists $\displaystyle \alpha >0$ such that :

$\displaystyle |f_n(x)-f(x)|=\frac{1}{x(1+nx^2)} \le \frac{1}{\alpha(1+n\alpha^2)}$

And when $\displaystyle n \to \infty$, $\displaystyle \frac{1}{\alpha(1+n\alpha^2)} \to 0$, that is to say that $\displaystyle \forall \varepsilon >0 ~,~ \exists N_0 \in \mathbb{N} ~,~ \forall n \in \mathbb{N} ~,~ n>N_0 \implies \forall x \in (0, \infty) ~,~ \frac{1}{\alpha(1+n\alpha^2)}< \varepsilon$

$\displaystyle \implies |f_n(x)-f(x)| \le \frac{1}{\alpha(1+n\alpha^2)}< \varepsilon.$

Which is what we're looking for...

I hope you understand & that it looks correct to you :/

Same thing for the following questions.