Answer
$$ \lim_{x\to 1} f(x) =1$$
Work Step by Step
Let $\epsilon\gt0$ and let $\delta=\min \left(1, \frac{\epsilon}{2}\right) .$ Then, whenever $|x-1|\lt\delta,$ it follows that $0\lt x\lt 2 .$ If $1\lt x\lt 2$ then $\min \left(x, x^{2}\right)=x$ and
$$
|f(x)-1|=|x-1|\lt\delta\lt \frac{\epsilon}{2}\lt \epsilon
$$
On the other hand, if $0\lt x\lt 1,$ then $\min \left(x, x^{2}\right)=x^{2},|x+1|\lt 2$ and
$$
|f(x)-1|=\left|x^{2}-1\right|=|x-1||x+1|\lt 2 \delta\lt \epsilon
$$
Hence
$$ \lim_{x\to 1} f(x) =1$$
