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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?

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    If $x = 1/2$, we would have $1/nx = 2/n > 1/n$. There is something unusual about the way the question is stated, because the target function is specifically defined on $[0, \infty)$ but then convergence is only claimed on $[1, \infty)$. So, unlike what the previous comment said, this does not really show a sequence of continuous functions converging uniformly to a discontinuous function; it really shows a sequence of functions that converges uniformly on a domain to a constant function on that domain. – 2017-02-28
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    Thank you both and yes, @CarlMummert it is unusually asked because I neglected to mention the first part of the question asked about uniform convergence on $[0,1]$, which obviously fails since we have $f_n$ cont for all n but $f$ discontinuous at x=0. Hence why f is defined on $[0, \infty)$ – 2017-02-28
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    @Carl Mummert - of course, you are right. Deleted my comment. – 2017-02-28
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    @user320014 What relevance has $x=0$ to a function defined on $[1,\infty)$? – 2017-02-28
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    @uniquesolution: the OP mentioned in a comment that there was a previous part to the exercise that looked at the interval $[0,1]$. – 2017-02-28
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    @user320014 You should clarify your question in the question-body, not in the comments. It does not make much sense, as it stands. – 2017-02-28

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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.

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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 $f(x)=1$?