$(f_n)_(n\geq2)$
$$f_n (x) = \begin{cases} 1-n^2x &\quad\text{for}\quad x \in [0,\frac{1}{n^2}]\\ 0 &\quad\text{for}\quad x \in (\frac{1}{n^2},\frac{2}{n^2}) \\ x - \frac{2}{n^2} &\quad\text{for}\quad x \in [\frac{2}{n^2},1]. \end{cases} $$
We just need to find the pointwise limit and i'm a little stuck.
I know that for $x=0$ we can do $f_n(0) = 1$ and hence $lim$ here is 1
But then I want to fix x between $0,1$ would i do this in separate cases. One between $(0,\frac{1}{n^2})$ and $[\frac{1}{n^2},1]$
I'm kind of stuck if anyone could show a working to this, it'd be really helpful
If $x>0$, then $\frac2{n^2}<x$ if $n\gg1$ and then you have $f_n(x)=x-\frac2{n^2}$. Since $\lim_{n\to\infty} x-\frac2{n^2}=x$, you have $\lim_{n\to\infty}f_n(x)=x$.