Determine the values of x for any holes in the graph of: 
f(x)= (x+3)/x^2+5x+6.

A hole occurs in a rational function f(x) at x=a if

f(a) is not defined at x=a due to numerator and denominator
becoming 0, and

lim f(x) = lim f(x) = lim f(x) = some finite number
x->a-      x->a+      x->a 

f(x) = (x+3)/[(x+3)(x+2)]  

We cannot cancel (x+3)'s except when x is not equal to -3,
for f(x) becomes the meaningless "0/0" and is not defined 
when x=-3.

However for every other number besides x=-3, 
function f(x) is identical with g(x) = 1/(x+2) because
the (x+3)'s can be canceled when (x+3) is not 0.

So

lim f(x) = g(-3) = 1/(-3+2) = -1
x->-3

So there is a hole in the curve at x = -3.

A hole in the curve is often called a "removable discontinuity"
because the hole could be plugged up by defining f(-3) as -1.

Edwin