Let $f(x) = \frac{nx}{1+nx}, x\in [0, \infty).$
- Find $\lim _{n \to \infty}f_n(x) = f(x).$
When $x = 0, f_n$ converges to $0$.
When $x\in(0,\infty)$, I claim that $\lim_{n\to\infty} f_n(x)=1.$
Proof: Let $\varepsilon>0$. Let $M\in N$ such that $M>\frac {1}{\varepsilon x}$. Then, for $n\ge M, |\frac {nx}{1+nx}-1| = |\frac {-1}{1+nx}|<\frac {1}{Mx} <\varepsilon.$
- Does $f_n \to f$ uniformly on $[0,1]$? Justify your answer.
Yes. Let $M > \frac 1{\varepsilon}.$ Then, for $n\ge M$, $|\frac {nx}{1+nx}-1|=|\frac {-1}{1+nx}|\le \frac {1}{1+n} <\frac 1n\le \frac 1M<\varepsilon $ for all $x\in(0,1].$ But, here, when $x=0,$ $f_n$ converges to $0$. So, it seems that $f_n$ still depends on whether $x= 0 $ or otherwise. How can we justify this?
- Does $f_n \to f$ uniformly on $[0,\infty)$? Justify your answer.
"No" because $|\frac {-1}{1+nx}|\not \le \frac {1}{1+n}$ (it now depends on whether $x\le 1 $ or $x>1$). But, I don't know how to prove formally by using negation of the definition. I appreciate any hint.
Thank you in advance.
The sequence $(f_n)_{n\in\mathbb N}$ doesn't converge uniformly to $f$ on $[0,1]$ (or on $[0,+\infty)$), because when a sequence of continuous functions converges uniformly to a function $f$, then $f$ is continuous too. But your function $f$ isn't continuous.