How do you find the limit of (3x + 9)/(x^2 - 9)3x+9x29 as x approaches -3?

2 Answers
Jul 8, 2018

-1/212

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

-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