Problem in finding limit function of $f_n(x)=1-(1-x^2)^{n}$. Its answer is $f(x) =
\begin{cases}
0,  & \text{when $x=0$ } \\
1, & \text{when $0<\vert x \vert< \sqrt 2$}
\end{cases}$.
I'm not getting how second line of $f(x)$ came?
Apologies if the post is too basic, but I've invested a great time in understanding this but of no use.
Any hints are welcome!

Assuming you want the limit as $n\to\infty$, then you need $|1-x^2|<1$. This is the open interval $(0,\sqrt 2)$. So
$$
\lim_{n\to\infty}f_n(x) =
\begin{cases}
0,  & \text{when $x=0$ } \\
1, & \text{when } 0<x< \sqrt 2.
\end{cases}
$$
For $x\geq \sqrt 2$ or $x<0$ the limit does not exist, since $|1-x^2|>1$.
