use properties of limits to find the indicated limit. it may be necessary to rewrite the expression before…

use properties of limits to find the indicated limit. it may be necessary to rewrite the expression before limit properties can be applied.\n lim_{x \to 3} \frac{x^{2}+5x - 24}{x^{2}-9} \nselect the correct choice below and, if necessary, fill in the answer box to complete your choice.\na. lim_{x \to 3} \frac{x^{2}+5x - 24}{x^{2}-9}=\frac{11}{6} (type an integer or a simplified fraction.)\nb. the limit does not exist and is neither (infty) nor (-infty).

use properties of limits to find the indicated limit. it may be necessary to rewrite the expression before limit properties can be applied.\n lim_{x \to 3} \frac{x^{2}+5x - 24}{x^{2}-9} \nselect the correct choice below and, if necessary, fill in the answer box to complete your choice.\na. lim_{x \to 3} \frac{x^{2}+5x - 24}{x^{2}-9}=\frac{11}{6} (type an integer or a simplified fraction.)\nb. the limit does not exist and is neither (infty) nor (-infty).

Answer

Explanation:

Step1: Factor the numerator and denominator

The numerator $x^{2}+5x - 24=(x + 8)(x-3)$ and the denominator $x^{2}-9=(x + 3)(x - 3)$. So the limit becomes $\lim_{x\rightarrow3}\frac{(x + 8)(x - 3)}{(x + 3)(x - 3)}$.

Step2: Cancel out the common factor

Since $x\neq3$ when taking the limit, we can cancel out the $(x - 3)$ terms. The limit simplifies to $\lim_{x\rightarrow3}\frac{x + 8}{x+3}$.

Step3: Substitute the value of x

Substitute $x = 3$ into $\frac{x + 8}{x+3}$, we get $\frac{3+8}{3 + 3}=\frac{11}{6}$.

Answer:

A. $\lim_{x\rightarrow3}\frac{x^{2}+5x - 24}{x^{2}-9}=\frac{11}{6}$