Answer
We can define $f(1)=\frac{3}{2}$ that extends $f(s)$ to be continuous at $s=1$.
Work Step by Step
$$f(s)=\frac{s^3-1}{s^2-1}$$
Currently, $f(s)$ is not defined at $s=1$, so $f(s)$ is not continuous at $s=1$ as well.
For $f(s)$ to be continuous at $s=1$, we need to extend $f(s)$ to include a value of $f(1)$ so that $f(1)=\lim_{s\to1}f(s)$
So we need to calculate $\lim_{s\to1}f(s)$ first:
$$\lim_{s\to1}f(s)=\lim_{s\to1}\frac{s^3-1}{s^2-1}=\lim_{s\to1}\frac{(s-1)(s^2+s+1)}{(s-1)(s+1)}=\lim_{s\to1}\frac{s^2+s+1}{s+1}$$
$$\lim_{s\to1}f(s)=\frac{1^2+1+1}{1+1}=\frac{3}{2}$$
So we can define $f(1)=\frac{3}{2}$ that extends $f(s)$ to be continuous at $s=1$.
