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

2 Answers
Jul 8, 2018

Answer:

#-1/2#

Explanation:

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#

Jul 8, 2018

Answer:

#-1/2#

Explanation:

#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#