When we deal with limits at infinity, it's always helpful to factor out an x, or an x^2, or whatever power of x simplifies the problem. For this one, let's factor out an x^2 from the numerator and an x from the denominator:
lim_(x->-oo)(sqrt(x^2-9))/(2x-6)=(sqrt((x^2)(1-9/(x^2))))/(x(2-6/x))
=(sqrt(x^2)sqrt(1-9/(x^2)))/(x(2-6/x))
Here's where it begins to get interesting. For x>0, sqrt(x^2) is positive; however, for x<0, sqrt(x^2) is negative. In mathematical terms:
sqrt(x^2)=abs(x) for x>0
sqrt(x^2)=-x for x<0
Since we're dealing with a limit at negative infinity, sqrt(x^2) becomes -x:
=(-xsqrt(1-9/(x^2)))/(x(2-6/x))
=(-sqrt(1-9/(x^2)))/(2-6/x)
Now we can see the beauty of this method: we have a 9/x^2 and 6/x, both of which will go to 0 as x goes to negative infinity:
lim_(x->-oo)=(-sqrt(1-0))/(2-0)
lim_(x->-oo)=-1/2
