Question

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

Answer 1

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:
`(x^2-9)/(x+3)=((x+3)(x-3))/(x+3)=(cancel((x+3))(x-3))/cancel((x+3))=`
now you have: `f(x)=x-3`
so that now you have:
`lim_(x->-3)(x^2-9)/(x+3)=lim_(x->-3)(x-3)=-6=f(-3)`
Your `x=-3` is a discontinuity that can be removed"!

so that basically `f(-3)=-6`

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