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

1 Answer
Gió
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:
(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]}

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