How do you prove that the limit of # (x²-9) / (x²+5x+6)=0 # as x approaches -3 using the epsilon delta proof?

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Answer 1 · 2017-01-09T10:33:37

The question is ill posed as:
#lim_(x->-3) (x^2-9)/(x^2+5x+6) !=0#

First we note that:

#x^2-9 = (x-3)(x+3)#

#x^2+5x+6 = (x+3)(x+2)#

So the function can be simplified as:

#frac (x^2-9) (x^2+5x+6) = frac ((x-3)(x+3)) ((x+3)(x+2)) = frac (x-3)(x+2)#

In this form the function is continuous for #x=-3#, so:

#lim_(x->-3) (x^2-9)/(x^2+5x+6) =lim_(x->-3) frac (x-3)(x+2) = (-3-3)/(-3+2) =(-6)/(-1) = 6#