Answer
The function has a discontinuity at $\pm 1$. Moreover the discontinuity is infinite because $\lim\limits_{t \to \pm1}=\infty $ .
Work Step by Step
Since $$ t^2-1=0 \Longrightarrow (t-1)(t+1)=0\Longrightarrow t=\pm1$$
then the function has a discontinuity at $\pm 1$. Moreover, the discontinuity is infinite because $\lim\limits_{t \to \pm1}=\infty $ .
