If $f_n(x)$ uniformly converges to a bounded function $f(x)$, then $f_n(x)$ is bounded for all $n \in \Bbb{N^+}$. Is this true?
I construct a counterexample as follows:
$f_n(x)= \begin{cases}1/x &n=1 \\ x+\frac{1}{n} &n\ge2 \end{cases}$, $\quad x \in(0,1)$
$f_n(x)$ uniformly converges to $x$, whereas $f_1(x)$ is unbounded on $(0,1)$.
My questions:
(1) Is this example correct?
(2) Can it be claimed that there exists $N$ such that $f_n(x)$ is bounded for every $n \ge N$ if $f_n(x)$ uniformly converges to a bounded function $f(x)$?