Answer
See explanations.
Work Step by Step
Step 1. We can identify the quantities as: $f(x)=\frac{1}{x},c=1,L=1$.
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 $|\frac{1}{x}-1|\lt\epsilon$
Step 3. The last inequality gives $-\epsilon\lt \frac{1}{x}-1\lt\epsilon$ or $1-\epsilon\lt \frac{1}{x}\lt1+\epsilon$
Step 4. Assume $\epsilon\lt1$, and $x\gt0$; we can get $\frac{1}{1+\epsilon}\lt x\lt \frac{1}{1-\epsilon}$
Step 5. From the requirement for the value $\delta$ in step 2, we have $-\delta\lt x-1\lt\delta$ or $1-\delta\lt x\lt1+\delta$
Step 6. Compare the results from steps 4 and 5, let $1-\delta=\frac{1}{1+\epsilon}$, we get $\delta=1-\frac{1}{1+\epsilon}=\frac{\epsilon}{1+\epsilon}$. Let $1+\delta=\frac{1}{1-\epsilon}$, we get $\delta=\frac{1}{1-\epsilon}-1=\frac{\epsilon}{1-\epsilon}$. Thus, we can set $\delta$ as the smaller value of $\frac{\epsilon}{1+\epsilon}$ and $\frac{\epsilon}{1-\epsilon}$
Step 7. For $\epsilon\geq1$, choose $\delta$ to be the smaller of $1$ and $\frac{\epsilon}{1+\epsilon}$
Step 8. With the setting of the $\delta$ value, we can go backwards through the above steps to reach a conclusion that for all $x$ when $0\lt |x-1|\lt\delta$, we have $|\frac{1}{x}-1|\lt\epsilon$ which proves the limit statement.
