In this particular case points #x=0# and #x=5# are exactly where the denominator is #0#, while a numerator is not.

Therefore, the function is not only undefined at these two points, but goes to infinity (#+oo# or #-oo#) as its argument approaches these values. In other words, #x=0# and #x=5# are *asymptotes* and function behavior around these two points is *asymptotic*.

At #x=0# the numerator equals to #3#.

As #x# approaches #0# from the left, the function is always positive (since #x<0# and #x-5<0#) and, therefore, it tends to #+oo#.

As #x# approaches #0# from the right, the function is always negative (since #x>0# and #x-5<0#) and, therefore, it tends to #-oo#.

At #x=5# the numerator equals to #8#.

As #x# approaches #5# from the left, the function is always negative (since #x>0# and #x-5<0#) and, therefore, it tends to #-oo#.

As #x# approaches #5# from the right, the function is always positive (since #x>0# and #x-5>0#) and, therefore, it tends to #+oo#.

Here is the graph of this function:

graph{(x+3)/(x(x-5)) [-5, 8, -5, 5]}