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

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Answer 1 · 2018-07-08T06:04:13

#-1/2#

Given: #lim x-> -3 " of " f(x) = (3x+ 9)/(x^2 - 9)#

First factor both the numerator and denominator. Note that the denominator is the difference of squares, #(a^2 - b^2) = (a-b)(a+b)#

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

Now take the limit at #x = -3#

#lim x-> -3 " of " f(-3) = 3/(-3 - 3) = 3/-6 = -1/2#

Answer 2 · 2018-07-08T10:59:57

#-1/2#

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

#=lim_(xto-3)(3cancel((x+3)))/(cancel((x+3))(x-3))#

#=lim_(xto-3)3/(x-3)=3/(-6)=-1/2#