Answer
See explanations.
Work Step by Step
Step 1. We only need to use the first equation of the piece-wise function as when $x\to1$, $x\ne1$. Thus we can identify the quantities as: $f(x)=x^2, c=1, L=1$.
Step 2. For a given small value $\epsilon\gt0$, we need to find a value $\delta\gt0$ so that for all $x$ when $0\lt |x-1|\lt \delta$, we have $|x^2-1|\lt\epsilon$
Step 3. The last inequality from the last step gives $-\epsilon\lt x^2-1\lt\epsilon$ or $1-\epsilon\lt x^2\lt1+\epsilon$
Step 4. Assume $\epsilon\lt1, x\gt0$; we can get $\sqrt {1-\epsilon}\lt x\lt\sqrt {1+\epsilon}$ and $1-\sqrt {1-\epsilon}\lt|x-1|\lt \sqrt {1+\epsilon}-1$
Step 5. Compare the result from step 4 with the requirement for the value $\delta$ in step 2; we can set $\delta$ as the smaller value of $1-\sqrt {1-\epsilon}$ and $ \sqrt {1+\epsilon}-1$
Step 6. For $\epsilon\geq1$, choose $\delta$ to be the smaller of $1$ and $\sqrt {1+\epsilon}-1$
Step 7. With the setting of the $\delta$ value, we can go backwards through the steps of 4,3,2 to reach a conclusion that for all $x$ when $0\lt |x-1|\lt \delta$, we have $|x^2-1|\lt\epsilon$ which proves the limit statement.
