Question

1. - / 1 points evaluate the limit, if it exists. (if an answer does not exist, enter dne.) $lim_{u

ightarrow - 5}\frac{u + 5}{u^{3}+125}$

Explanation:

Step1: Factor the denominator

We know that $a^{3}+b^{3}=(a + b)(a^{2}-ab + b^{2})$. Here, $a = u$ and $b = 5$, so $u^{3}+125=(u + 5)(u^{2}-5u + 25)$.
So the limit becomes $\lim_{u
ightarrow - 5}\frac{u + 5}{(u + 5)(u^{2}-5u + 25)}$.

Step2: Simplify the function

Cancel out the common factor $(u + 5)$ (since $u
eq - 5$ when taking the limit), we get $\lim_{u
ightarrow - 5}\frac{1}{u^{2}-5u + 25}$.

Step3: Substitute the value of $u$

Substitute $u=-5$ into $\frac{1}{u^{2}-5u + 25}$, we have $\frac{1}{(-5)^{2}-5\times(-5)+25}=\frac{1}{25 + 25+25}=\frac{1}{75}$.

Answer:

$\frac{1}{75}$