In order for a function #f(x)# to be continuous at a given #x#-value #a#, the following condition must be satisfied:
#color(gray)([1]) lim_(x->a)f(x)=f(a)#
What this is saying is that, as #x# gets closer to #a#, #f(x)# should also get closer to #f(a)#.
For the given function #f(x)#, the limit on the left-hand side of #color(gray)([1])# will evaluate correctly. You'll end up with #lim_(x->3)(x^2-9)/(x-3)=6.# However, the right-hand side of #color(gray)([1])# presents a problem: what is #f(x)# when #x=3?# The answer is, it is not defined, because at that point, we have #f(x)# "equal" to #0/0#:
#f(3)=(3^2-9)/(3-3)=(9-9)/0=0/0#
And this "value" of #0/0# is indeterminate.
Thus, the function "breaks" at #x=3#, and so, because there is no #f(3)#, it is not possible to say #lim_(x->3)f(x)=f(3).# Thus, #f(x)# is not continuous at #x=3.#
Bonus:
Here's a tip to determine where some functions will be discontinuous. Any #x#-value that makes any denominator in the function equal to #0# is a point of discontinuity. So for the function above, #x-3=0# when #x=3#, and so the function will be discontinuous at #x=3.#
The only exception to this is if the function is piecewise defined at potential "breaking" points, like this:
#f^star(x)={((x^2-9)/(x-3),",",x!=3),(6,",",x=3):}#
In this case, since #f^star(x)# is defined at #x=3#, we have #lim_(x->3)f^star(x)=f^star(3),# so #f^star(x)# is continuous at #x=3# (and everywhere).
