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#
