What does it mean when it says: "non-removable discontinuity at x=4"?

Here is the definition I am familiar with.

Function #f# has a discontinuity at #a# if #f# is defined in an open interval containing #a# except possibly at #x=a#, and #f# is not continuous at #a#.

(This allows us to avoid saying, for example that #sqrtx# has a discontinuity at #x=-5#, which is good. But, the function #f(x) = 1# with domain all irrational numbers is not continuous at #3#, but is also not discontinuous at #3#, because is it not defined on an open interval containing #3#.)

#f# has a removable discontinuity at #a# if #lim_(xrarra) f(x)# exists. (Remember that writing #lim_(xrarra)f(x)=oo# is a way of explaining why the limit Does Not Exist.)

#f# has a non-removable discontinuity at #a# if #f# has a discontinuity at #a# and #lim_(xrarra) f(x)# does not exist.

In the most familiar functions:

Rational and trigonometric functions have non-removable discontinuities at their vertical asymptotes. (Holes are removable.)

Piecewise-defined functions can have jump discontinuities, which are non-removable. (Holes are removable.)

The Greatest integer function (a.k.a. the Floor function) has a non-removable discontinuity at every integer.