Answer
The limit $ \lim _{t \rightarrow 9} \frac{t-6}{\sqrt{t}-3} $ does not exist. The one-sided limits are infinite.
Work Step by Step
By substitution, we get
$$ \lim _{t \rightarrow 9} \frac{t-6}{\sqrt{t}-3}=\frac{9-6}{3-3}=\frac{3}{0} $$
which means that the limit does not exist. The one-sided limits can be calculated as follows
$$ \lim _{t \rightarrow 9^+} \frac{t-6}{\sqrt{t}-3}=\frac{9-6}{3^+-3}=\infty $$
and
$$ \lim _{t \rightarrow 9^-} \frac{t-6}{\sqrt{t}-3}=\frac{9-6}{3^--3}=-\infty $$
hence the limit $ \lim _{t \rightarrow 9} \frac{t-6}{\sqrt{t}-3} $ does not exist.
