question 22 (1 point) find the limit algebraically. lim x→5 x2 - 2x - 15 / x + 3 a 0 b 5 c -8 d does not exist

Explanation:

Step1: Factor the numerator

Factor $x^{2}-2x - 15$ as $(x - 5)(x+3)$. So the function becomes $\lim_{x\rightarrow5}\frac{(x - 5)(x + 3)}{x+3}$.

Step2: Simplify the function

Cancel out the common factor $(x + 3)$ (since $x\neq - 3$ when $x\rightarrow5$). The function simplifies to $\lim_{x\rightarrow5}(x - 5)$.

Step3: Evaluate the limit

Substitute $x = 5$ into $x-5$. We get $5-5=0$.

Answer:

A. 0