How do you find the limit of $\frac{3 x + 9}{x^{2} - 9}$ $(3x + 9)/(x^2 - 9)$ as x approaches -3?

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Answer 1 · 2018-07-08T06:04:13

$- \frac{1}{2}$ $-1/2$

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

$-1/2$

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

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

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

How do you find the limit of (3x + 9)/(x^2 - 9)3x+9x2−9(3x + 9)/(x^2 - 9) as x approaches -3?