How do you find the limit of #(x^2+x-6)/(x^2-9)# as #x->-3#?

Question

8272 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-08T03:30:08

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

Factorize the numerator and denominator:

#x^2+x-6 = 0#

#x= frac (-1+-sqrt(1+24)) 2 = frac (-1+-5) 2#

#x_1 = -3#
#x_2=2#

so:

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

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

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