Is $\frac{nx}{1+nx^2}$ a Cauchy sequence in $C([0,1])$?

Question

Is the sequence of functions $f_n(x)=\frac{nx}{1+nx^2}$ a Cauchy sequence in $C([0,1])$?

I'm a little lost as to how to go about this. I thought I could just check $|f_n(x)-f_m(x)|$ and show that it is/isn't $<\epsilon$, for $\epsilon$ small, but I confused myself somewhere. Any help/hints would be greatly appreciated.

@NormalHuman Thank you for catching that. I think this is under real analysis? — Desperate Fluffy, Oct 27 '15 at 22:00
Could you show use what you have done to confuse yourself? Maybe we can enlighten you on some pesky detail/miscalculation. — Hetebrij, Oct 27 '15 at 22:04
Hint: $f_n(1/\sqrt{n})=\frac{\sqrt{n}}{2}$. I came up with this by determining which term is dominant in the denominator for various values of $n$ and $x$; the $1$ term is dominant for $x \ll 1/\sqrt{n}$ while the other term is dominant for $x \gg 1/\sqrt{n}$. This is often a good way to prove that the maximum of a function is big without explicitly computing it. (Of course in this simple case you can in fact explicitly compute it, and it turns out to be exactly at $1/\sqrt{n}$, but that is unusual.) — Ian, Oct 27 '15 at 22:04
@Ian I'm sorry, but I'm not sure what to do with that. A Cauchy sequence is one that satisfies $|f_n-f_{n+1}|<\epsilon$, right? If so, I'm still struggling with putting that in terms of this sequence. If not...I'm completely and utterly lost. — Desperate Fluffy, Oct 27 '15 at 22:24
@Did I'm sorry, but I don't understand the question. I wasn't given a distance. — Desperate Fluffy, Oct 27 '15 at 22:44
@DesperateFluffy A Cauchy sequence is bounded, but what I wrote shows that your $f_n$ is not bounded in the uniform metric. (Note that your question makes no sense without reference to a metric.) — Ian, Oct 27 '15 at 22:48
@DesperateFluffy How do you define a sequence $(x_n)$ with values in some space $E$ to be Cauchy, in general? — Did, Oct 27 '15 at 23:02

score 2 · Accepted Answer · answered Oct 27 '15 at 22:46

No, it is not a Cauchy sequence in the space $C[0,1]$ with the supremum norm. If it were, then as the space is complete, it would have converged uniformly to a continuous function, and in particular, it would have converged point-wise to a continuous function, but clearly for $x\neq 0$,

$$\lim_{n\to\infty}\frac{nx}{1+nx^2}=\frac{1}{x},$$ and there is no continuous function on $[0,1]$ which is equal to $1/x$ for $x\neq 0$.

Is $\frac{nx}{1+nx^2}$ a Cauchy sequence in $C([0,1])$?

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