Answer
$$\lim_{x \to 2} (1-x^2)=-3$$
Work Step by Step
$$\lim_{x \to 2} (1-x^2)=1-2^2=-3$$Now, we want to prove this limit by using $\varepsilon - \delta$ definition; that is, we must show that for each $\varepsilon >0$, there exists a $\delta >0$ such that $|(1-x^2)-(-3)|< \varepsilon$ whenever $|x-2|< \delta$.
Now, we have$$|(1-x^2)-(-3)|=|4-x^2|=|x^2-4|=|x-2||x+2|.$$For all $x$ in the interval $(1,3)$, $|x+2|<5$. So, letting $\delta$ be the minimum of $1$ and $\frac{\varepsilon }{5}$, we conclude that$$|x-2|< \delta =\min \{ 1, \frac{\varepsilon }{5} \} \quad \Rightarrow \quad |(1-x^2)-(-3)|=|x-2||x+2|
