Question

hw5 the limit laws (target l4; §2.3)
score: 7/13 answered: 7/13
question 8
evaluate the limit:
\\(\lim_{x\to - 8}\frac{-2x - 16}{x^{2}+9x + 8}=\\)
question help: video message instructor

Explanation:

Step1: Factor the numerator and denominator

First, factor the numerator $-2x - 16=-2(x + 8)$ and the denominator $x^{2}+9x + 8=(x + 8)(x+1)$.
So the function becomes $\frac{-2(x + 8)}{(x + 8)(x + 1)}$.

Step2: Simplify the function

Cancel out the common factor $(x + 8)$ (since $x\to - 8$ but $x
eq - 8$ when taking the limit), we get $\frac{-2}{x + 1}$.

Step3: Substitute the value of $x$

Substitute $x=-8$ into $\frac{-2}{x + 1}$, we have $\frac{-2}{-8 + 1}=\frac{-2}{-7}=\frac{2}{7}$.

Answer:

$\frac{2}{7}$