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

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Answer 1 · 2017-01-31T00:43:41

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

Simplify the rational function noting that:

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

#x^2+2x-3 = (x-1)(x+3)#

so:

#(x^2-9)/(x^2+2x-3)= ((x-3)(x+3))/((x-1)(x+3)) =(x-3)/(x-1)#

We have thus eliminated the discontinuity for #x=-3# and we have:

#lim_(x->-3) (x^2-9)/((x-1)(x+3)) = lim_(x->-3)(x-3)/(x-1) = (-6)/-4= 3/2#