How do you evaluate the limit #(x^2-9)/(x^3-27)# as x approaches #3#?

Question

Isabella Ackerman · Accepted Answer

#lim_(xrarr3) (x^2-9)/(x^3-27) =color(green)(2/9)# Two general equations to remember: #color(white)("XXX")a^2-b^2 = (a-b)(a+b)# #color(white)("XXX")a^3-b^3=(a-b)(a^2+ab+b^2)# Therefore #color(white)("XXX")(x^2-9)/(x^3-27) = ((x-3)(x+3))/((x-3)(x^2+3x+9))# As long as #x!=3# (which is true as long as #x# is only approaching #3#) the factors #(x-3)# can be cancelled out of the numerator an denominator. So #color(white)("XXX")lim_(xrarr3) (x^2-9)/(x^3-27)# #color(white)("XXXXXX") = lim_(xrarr3) (x+3)/(x^2+3x+9)# #color(white)("XXXXXX")=(3+3)/(9+9+9)# #color(white)("XXXXXX")=6/27# #color(white)("XXXXXX")=2/9#

Cameron Alexander · Answer

undefined To evaluate the limit (x^2-9)/(x^3-27) as x approaches 3, we can substitute the value of x into the expression. Plugging in x=3, we get (3^2-9)/(3^3-27). Simplifying this expression gives us (0)/(0). This is an indeterminate form, so we need to further simplify. Factoring the numerator and denominator, we have (3-3)(3+3)/(3-3)(3^2+3*3+3^2). Canceling out the common factors of (3-3), we are left with (3+3)/(3^2+3*3+3^2). Evaluating this expression gives us 6/27, which simplifies to 2/9. Therefore, the limit of (x^2-9)/(x^3-27) as x approaches 3 is 2/9.