# What is the continuity of f(t) = 3 - sqrt(9-t^2)?

May 1, 2015

$f \left(t\right) = 3 - \sqrt{9 - {t}^{2}}$ has domain $\left[- 3 , 3\right]$

For $a$ in $\left(- 3 , 3\right)$, ${\lim}_{t \rightarrow a} f \left(t\right) = f \left(a\right)$ because

${\lim}_{t \rightarrow a} \left(3 - \sqrt{9 - {t}^{2}}\right) = 3 - {\lim}_{t \rightarrow a} \sqrt{9 - {t}^{2}}$

= 3-sqrt(lim_(trarra) (9-t^2)) = 3-sqrt(9-lim_(trarra) t^2))

$= 3 - \sqrt{9 - {a}^{2}} = f \left(a\right)$

So $f$ is continuous on $\left(- 3 , 3\right)$.

Similar reasoning will show that

${\lim}_{t \rightarrow - {3}^{+}} f \left(t\right) = f \left(- 3\right)$ and

${\lim}_{t \rightarrow {3}^{-}} f \left(t\right) = f \left(3\right)$

So $f$ is continuous on $\left[- 3 , 3\right]$.