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

1 Answer
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]}