let f(x) = (2x² - 9x - 5)/(x² + x - 30). find the indicated quantities, if they exist. (a) lim(x→5) f(x) (b)…

let f(x) = (2x² - 9x - 5)/(x² + x - 30). find the indicated quantities, if they exist. (a) lim(x→5) f(x) (b) lim(x→0) f(x) (c) lim(x→1) f(x) (a) select the correct choice below and, if necessary, fill in the answer box to complete your choice. o a. lim(x→5) (2x² - 9x - 5)/(x² + x - 30) = (type an integer or a simplified fraction.) o b. the limit does not exist. (b) select the correct choice below and, if necessary, fill in the answer box to complete your choice. o a. lim(x→0) (2x² - 9x - 5)/(x² + x - 30) = (type an integer or a simplified fraction.) o b. the limit does not exist. (c) select the correct choice below and, if necessary, fill in the answer box to complete your choice. o a. lim(x→1) (2x² - 9x - 5)/(x² + x - 30) = (type an integer or a simplified fraction.) o b. the limit does not exist.

let f(x) = (2x² - 9x - 5)/(x² + x - 30). find the indicated quantities, if they exist. (a) lim(x→5) f(x) (b) lim(x→0) f(x) (c) lim(x→1) f(x) (a) select the correct choice below and, if necessary, fill in the answer box to complete your choice. o a. lim(x→5) (2x² - 9x - 5)/(x² + x - 30) = (type an integer or a simplified fraction.) o b. the limit does not exist. (b) select the correct choice below and, if necessary, fill in the answer box to complete your choice. o a. lim(x→0) (2x² - 9x - 5)/(x² + x - 30) = (type an integer or a simplified fraction.) o b. the limit does not exist. (c) select the correct choice below and, if necessary, fill in the answer box to complete your choice. o a. lim(x→1) (2x² - 9x - 5)/(x² + x - 30) = (type an integer or a simplified fraction.) o b. the limit does not exist.

Answer

Explanation:

Step1: Factor the numerator and denominator

Factor $2x^{2}-9x - 5=(2x + 1)(x - 5)$ and $x^{2}+x - 30=(x + 6)(x - 5)$. So $f(x)=\frac{(2x + 1)(x - 5)}{(x + 6)(x - 5)}=\frac{2x+1}{x + 6},x\neq5$.

Step2: Calculate $\lim_{x\rightarrow5}f(x)$

Substitute $x = 5$ into $\frac{2x+1}{x + 6}$. We get $\lim_{x\rightarrow5}\frac{2x+1}{x + 6}=\frac{2\times5+1}{5 + 6}=\frac{10 + 1}{11}=\frac{11}{11}=1$.

Step3: Calculate $\lim_{x\rightarrow0}f(x)$

Substitute $x = 0$ into $\frac{2x+1}{x + 6}$. We get $\lim_{x\rightarrow0}\frac{2x+1}{x + 6}=\frac{2\times0+1}{0 + 6}=\frac{1}{6}$.

Step4: Calculate $\lim_{x\rightarrow1}f(x)$

Substitute $x = 1$ into $\frac{2x+1}{x + 6}$. We get $\lim_{x\rightarrow1}\frac{2x+1}{x + 6}=\frac{2\times1+1}{1 + 6}=\frac{2 + 1}{7}=\frac{3}{7}$.

Answer:

(A) A. $\lim_{x\rightarrow5}\frac{2x^{2}-9x - 5}{x^{2}+x - 30}=1$ (B) A. $\lim_{x\rightarrow0}\frac{2x^{2}-9x - 5}{x^{2}+x - 30}=\frac{1}{6}$ (C) A. $\lim_{x\rightarrow1}\frac{2x^{2}-9x - 5}{x^{2}+x - 30}=\frac{3}{7}$