Question

Kim ((sqrt t^2 -9)-3)/t^2 as t tends to 0 on both sides​

Answer 1

The limit doesn't exist. Notice that $\sqrt{t^2-9}$ is defined only for $t^2\ge9$, or $|t|\ge3$, or $t\le-3$ and $t\ge3$. In other words, $\dfrac{\sqrt{t^2-9}-3}{t^2}$ doesn't exist for any $t$ between -3 and 3.