question 25 (1 point) find the limit algebraically. you must show work in order to receive credit! lim(x→…

Explanation:

Step1: Factor the numerator and denominator

The numerator $x^{2}-16$ is a difference - of - squares and can be factored as $(x + 4)(x - 4)$. The denominator $x^{3}+64$ is a sum of cubes and can be factored using the formula $a^{3}+b^{3}=(a + b)(a^{2}-ab + b^{2})$, where $a=x$ and $b = 4$, so $x^{3}+64=(x + 4)(x^{2}-4x + 16)$. So, $\lim_{x\rightarrow - 4}\frac{x^{2}-16}{x^{3}+64}=\lim_{x\rightarrow - 4}\frac{(x + 4)(x - 4)}{(x + 4)(x^{2}-4x + 16)}$.

Step2: Cancel out the common factor

Cancel out the common factor $(x + 4)$ (since $x\neq - 4$ when taking the limit), we get $\lim_{x\rightarrow - 4}\frac{x - 4}{x^{2}-4x + 16}$.

Step3: Substitute $x=-4$

Substitute $x=-4$ into $\frac{x - 4}{x^{2}-4x + 16}$, we have $\frac{-4-4}{(-4)^{2}-4\times(-4)+16}=\frac{-8}{16 + 16+16}=\frac{-8}{48}=-\frac{1}{6}$.

Answer:

$-\frac{1}{6}$