## Calculus (3rd Edition)

The limit $\lim _{t \rightarrow 9} \frac{t-6}{\sqrt{t}-3}$ does not exist. The one-sided limits are infinite.
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.