Answer
Set $f( 1 )=\displaystyle \frac{3}{2}$
Work Step by Step
$f $ is not defined for $s=\pm1$, and $f$ is continuous everywhere else.
We check if $\displaystyle \lim_{s\rightarrow 1}f(s)=L$ exists, and if it does, we can define the continuous extension of $f$ at $s=1$ as
$F(s)=\left\{\begin{array}{ll}
f(s) , & \text{if }s\neq 1\\
L, & \text{if }s=1
\end{array}\right.$
$L= \displaystyle \lim_{s\rightarrow 1}f(s)= \displaystyle \lim_{x\rightarrow 3}\frac{s^{3}-1}{s^{2}-1}\quad $
... recognize difference of cubes, difference of squares
$= \displaystyle \lim_{s\rightarrow 1}\frac{(s-1)(s^{2}+s+1)}{(s-1)(s+1)}\quad $...cancel common term
$= \displaystyle \lim_{s\rightarrow 1}\frac{s^{2}+s+1}{s+1}$
... evaluate directly,
$L=\displaystyle \frac{3}{2}$
The limit exits at $s=1,$ so, if we redefine $h$,
$f(s)=\left\{\begin{array}{ll}
\dfrac{s^{3}-1}{s^{2}-1} , & \text{if }s\neq 1 \\\\
3/2, & \text{if }s=1
\end{array}\right.$
it becomes continuous.
