# How to calculate $\liminf$ of the function $f(x)=1-x^2$

Let $$f:\mathbb{R} \longrightarrow \mathbb{R}$$ given by $$f(x)= 1-x^2, \; \forall \; x \in \mathbb{R}.$$ Question. How to evaluate $$\liminf_{|x|\rightarrow \infty} f(x)?$$

What I got to think about is $$\liminf_{|x|\rightarrow \infty} f(x)=\min\{c \in \mathbb{R} \; ; \; x_n \rightarrow \infty \: \text{and} \: f(x_n)=c\}.$$ It's true? Or is my reasoning wrong?

The function $$f(x)=1-x^2$$ is not bounded at infinity, that is, for all $$A>0$$ and $$k>0$$ there exists $$x_0 \in \mathbb{R}$$ such that $$x_0>A$$ we have $$|f(x_0)|>k$$ $$(\lim_{x \to \infty} 1-x^2=-\infty)$$.
Since $$\liminf_{x \rightarrow \infty} f(x)=\sup_{N>0}\inf_{x \geq N} f(x)$$ and for all $$N>0$$ we have $$\inf_{x \geq N} f(x)=-\infty$$ then $$\liminf_{x \rightarrow \infty}(1-x^2)=-\infty.$$
$$\lim_{|x| \rightarrow +\infty} f(x) = -\infty$$
so $$\liminf_{|x| \rightarrow +\infty} f(x) = \limsup_{|x| \rightarrow +\infty} f(x) =-\infty$$