so I have a function $f_n:[0,1] \to \mathbb R$ defined as $$ f_n(x) = $$\begin{cases} \tfrac{2n}{n-1}x & 0\le x\le\tfrac{n-1}{2n}\\ 2-\tfrac{2n}{n-1}x & \tfrac{n-1}{2n}\le x \le\tfrac{n-1}{n} \\ 0 & \tfrac{n-1}{n}\le x \le\ 1 \end{cases} and I'm trying to figure out the pointwise limit. my reasoning so far is if $ x=0 $,$\lim_{n\to \infty} f_n(0)=0$ for all n.
the next part is the part im having a little trouble understanding, and any clarification would be much appreciated.
can i pretty much ignore the middle term by saying if $x>0$, then let $x>\tfrac{N-1}{N}$ , then $f_n(x)=0$ for all $n>N$. so the pointwise limit of $f(x)$ is $0$.
If this is correct (which i feel it is based on examples i've seen) is the reason this is so because the intervals involving n become arbitrarily small as n gets very large and so we can essentially ignore them ?
hint
For $x\in (0,\frac {1}{2}) $,
using the fact that for great enough $n $, we have
$$0 <x <\frac {1}{2}-\frac {1}{2n}<\frac {1}{2} $$ thus for large enough $n $,
$$f_n (x)=\frac {2nx}{n-1} $$ and $$\lim_{n\to+\infty}f_n (x)=2x .$$
For $ \frac {1}{2} \le x <1$ and great $n $,
$$\frac {n-1}{2n}\le x\le 1-\frac {1}{n} $$ thus $$\lim_{n\to+\infty}f_n (x)=2-2x $$
finally, for $x=1 \;, f_n (1)=0$.