Is this proof correct for $f_n(x) = \frac{nx}{1+nx}$ converges uniformly to $f(x)=1$ on $[1,\infty)$

Question

My analysis professor provided a proof for this but it's not an epsilon-style proof (it simply considers the limit as $n \to \infty$ under the domain of $[1,\infty)$).

Is this $\epsilon$-style proof valid for these types of questions? I mean, it looks valid to me but it is 04:30 in the morning >.> I wonder since this is what would come to my mind during a midterm, not my professor's style of proof.

$Proof.\; Let$ $\epsilon>0.$ $Let$ $N \in \mathbb N \;s.t.\;N > \frac 1\epsilon.$

$Then \;\; n>N \implies |f_n(x)-f(x)|=|\frac{nx}{1+nx}-1|=|\frac{1}{1+nx}| <|\frac{1}{nx}|=\frac{1}{nx}\le \frac 1n < \epsilon.\;\square$

I got rid of the absolute value signs since n,x are positive and got rid of x in the second to last inequality since $x \ge 1$. Is this acceptable to do?

Answer 1

Yes, the proof is correct. As you pointed out, it is vital to know that $x \geq 1$ in order to make the inequalities work out.

Answer 2

Yes!

The proof you have provided is perfectly valid. I do have one question, though.

You want to find the limit of the sequence of functions on $[1,\infty)$. Even so, you have taken the limit function to be: $$f(x)=\begin{cases}0&x=0 \\ 1&x>0 \end{cases}$$ If you want convergence on $[1,\infty)$, why not simply take $f(x)=1$?