Answer
See explanations.
Work Step by Step
Step 1. To prove that the function is continuous at $c=1$, we need to show that the limit of the function exists at this point and equals to a value of $f(1)=1^2-7=-6$.
Step 2. For any small positive $\epsilon\gt0$, we need to find a corresponding value $\delta \gt0$ so that for all $x$ in the interval $|x-1|\lt\delta$, we have $|x^2-7+6|\lt\epsilon$ or $|x^2-1|\lt\epsilon$
Step 3. The last inequality can be written as $-\epsilon\lt x^2-1\lt\epsilon$ or $1-\epsilon\lt x^2\lt\epsilon+1$
Step 4. Assuming $\epsilon\lt1$, we have $\sqrt{1-\epsilon}\lt x\lt \sqrt{1+\epsilon}$ and $\sqrt{1-\epsilon}-1\lt x-1\lt \sqrt{1+\epsilon}-1$.
Step 5 Choose $\delta$ to be the smaller of $|\sqrt{1-\epsilon}-1|$ 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 chosen $\delta$ value, working the above steps backwards, we have for all $x$ in the interval $|x-1|\lt\delta$, $|x^2-1|\lt\epsilon$, which proves that $\lim_{x\to1}f(x)=-6=f(1)$
Step 8. We conclude that the function is continuous at $c=1$
