# If f(x)= (x^2-9)/(x+3) is continuous at x= -3, then what is f(-3)?

May 27, 2015

Apparently your function is not continuous at $x = - 3$ because if you use this value you'd get a division by zero that cannot be performed.

But, if you write:
$\frac{{x}^{2} - 9}{x + 3} = \frac{\left(x + 3\right) \left(x - 3\right)}{x + 3} = \frac{\cancel{\left(x + 3\right)} \left(x - 3\right)}{\cancel{\left(x + 3\right)}} =$
now you have: $f \left(x\right) = x - 3$
so that now you have:
${\lim}_{x \to - 3} \frac{{x}^{2} - 9}{x + 3} = {\lim}_{x \to - 3} \left(x - 3\right) = - 6 = f \left(- 3\right)$
Your $x = - 3$ is a discontinuity that can be removed"!

so that basically $f \left(- 3\right) = - 6$

Graphically:
graph{(x^2-9)/(x+3) [-10, 10, -5, 5]}