hw5 the limit laws (target l4; §2.3)\nscore: 7/13 answered: 7/13\nquestion 8\nevaluate the…

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

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

Answer

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 $x\neq - 8$ 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:

$\frac{2}{7}$