# Prove that the function $f_n(x)=\frac{x}{n}, x \in [0, \infty).$

There are three subquestions. (a) Find $f(x) = \lim f_n(x)$

The limit $\lim_{n \rightarrow \infty}f_n(x) = \lim_{n \rightarrow \infty} \frac{x}{n} = \lim_{n \rightarrow \infty}x * \lim_{n \rightarrow \infty} \frac{1}{n} = 0.$

(b) Determine whether $f_n \rightarrow f$ uniformly on $[0,1].$

Let $\epsilon >0$ and $N=\frac{1}{\epsilon}.$ Then, $\forall n > N = \frac{1}{\epsilon} \Rightarrow \epsilon > \frac{x}{n}.$

The question that I have problem on is the third one. The following is what I did but I am not sure if it is correct. (c) Determine whether $f_n \rightarrow f$ uniformly on $[0,\infty).$

For $\epsilon > 0$, consider the function when $x = \epsilon*n.$ Then, $\forall n > N \Rightarrow \left| \frac{x}{n} \right| = \left| \frac{\epsilon*n}{n} \right| = \epsilon,$ which implies that the sequence of function is not uniformly continuous on the gvien interval.

The reason I am confusing on the third one is because of the following theorem. Any comment??

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What is the theorem you mentioned in the last sentence? –  John Nov 11 '13 at 2:15
your reasoning is right, but the explaining needs to be clarified. in particular you cannot set $x=\epsilon*n$, and then quantify other $n$: it is the other way around. –  Denis Nov 11 '13 at 2:23
@John never mind about that theorem...thanks –  eChung00 Nov 11 '13 at 2:50
@dkuper What if I set $\epsilon$ equal to some value and prove it?? Is it ok?? –  eChung00 Nov 11 '13 at 2:51
Here is an intuitive way to think of it -- the larger x is the larger n has to be to drag x/n down towards 0. So the choice of N where for all n > N $|f_n| < \epsilon$ depends on x. So the convergence cannot be uniform if x is unbounded. Can you cast that into nice mathematical language? –  Betty Mock Nov 11 '13 at 3:45

$\exists\epsilon>0,\forall N,\exists x,\exists n\geq N,f_n(x)>\epsilon$.
It means that you can choose $\epsilon$ as a constant, and that $x$ is allowed to depend on $\epsilon$ and $N$.
So you made the right choice with $x=N*\epsilon$, but you can even fix $\epsilon$ to 1. Then taking $n=N+1$ works.