Question

hw 2.2
score: 3.75/8 answered: 4/8
question 5
evaluate the limit:
\\(\lim_{x \to -3} \frac{-4x - 12}{x^2 - 3x - 18}\\) =
question help: video
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Explanation:

Step1: Factor numerator and denominator

Factor numerator: $-4x - 12 = -4(x + 3)$
Factor denominator: $x^2 - 3x - 18 = (x - 6)(x + 3)$
So the expression becomes $\lim_{x \to -3} \frac{-4(x + 3)}{(x - 6)(x + 3)}$

Step2: Cancel common factor

Cancel $(x + 3)$ (since $x \to -3$, $x
eq -3$, so cancellation is valid):
$\lim_{x \to -3} \frac{-4}{x - 6}$

Step3: Substitute $x = -3$

Substitute $x = -3$ into $\frac{-4}{x - 6}$:
$\frac{-4}{-3 - 6} = \frac{-4}{-9} = \frac{4}{9}$

Answer:

$\frac{4}{9}$