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).