Let $\displaystyle f^{(n)}(x) := 2n $ when $\displaystyle x \in [1/2n, 1/n] $ and $\displaystyle f^{(n)}(x) := 0 $ for all other values of $\displaystyle x $. Why does $\displaystyle f^{(n)} $ converge pointwise to $\displaystyle f(x) := 0 $?
Pointwise convergence depends on both n and x.
So for any fixed x if it is out side of $\displaystyle [0,1]$
$\displaystyle f^{n}=0$ for all n
so we are okay.
Now for any $\displaystyle x \in [0,1]$
There exits an N such that for all n> N $\displaystyle f^{n}=0$
The Key point is that it depends both on x and n. I hope this helps
pay attention to your interval. $\displaystyle [1/2,1]$ is the BIGGEST it will be. for your example, $\displaystyle x = 100$, we have the interval $\displaystyle [1/200, 1/100]$. this is why TES considers the interval $\displaystyle (0,1]$, because $\displaystyle [1/(2n), 1/n] \to \{ 0 \}$ (note, 0 is never actually in the interval)
so as $\displaystyle n$ gets large, the interval for which $\displaystyle f^{(n)} \ne 0$ gets smaller and smaller...
I meant $\displaystyle (0,1]$
What we have is a sequnce of intervals.
When n=1 $\displaystyle [\frac{1}{2},1]$
n=2 $\displaystyle [\frac{1}{4},\frac{1}{2}]$
n=3 $\displaystyle [\frac{1}{6},\frac{1}{3}]$
n=4 $\displaystyle [\frac{1}{8},\frac{1}{4}]$
...
So eventally every x in $\displaystyle (0,1]$
Is not in $\displaystyle [\frac{1}{2n},\frac{1}{n}]$
so $\displaystyle f^{n}=0$