let f(x)=5x² - 29x - 6 / x² + 19x - 150. find the indicated quantities, if they exist. (a) lim x→6 f(x) (b)…

let f(x)=5x² - 29x - 6 / x² + 19x - 150. find the indicated quantities, if they exist. (a) lim x→6 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. a. lim x→6 5x² - 29x - 6 / x² + 19x - 150 = (type an integer or a simplified fraction.) b. the limit does not exist. (b) select the correct choice below and, if necessary, fill in the answer box to complete your choice. a. lim x→0 5x² - 29x - 6 / x² + 19x - 150 = (type an integer or a simplified fraction.) b. the limit does not exist. (c) select the correct choice below and, if necessary, fill in the answer box to complete your choice. a. lim x→1 5x² - 29x - 6 / x² + 19x - 150 = (type an integer or a simplified fraction.) b. the limit does not exist.

let f(x)=5x² - 29x - 6 / x² + 19x - 150. find the indicated quantities, if they exist. (a) lim x→6 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. a. lim x→6 5x² - 29x - 6 / x² + 19x - 150 = (type an integer or a simplified fraction.) b. the limit does not exist. (b) select the correct choice below and, if necessary, fill in the answer box to complete your choice. a. lim x→0 5x² - 29x - 6 / x² + 19x - 150 = (type an integer or a simplified fraction.) b. the limit does not exist. (c) select the correct choice below and, if necessary, fill in the answer box to complete your choice. a. lim x→1 5x² - 29x - 6 / x² + 19x - 150 = (type an integer or a simplified fraction.) b. the limit does not exist.

Answer

Explanation:

Step1: Factor the numerator and denominator

The numerator $5x^{2}-29x - 6=(5x + 1)(x - 6)$. The denominator $x^{2}+19x - 150=(x + 25)(x - 6)$. So $f(x)=\frac{(5x + 1)(x - 6)}{(x + 25)(x - 6)}=\frac{5x+1}{x + 25},x\neq6$.

Step2: Calculate $\lim_{x\rightarrow6}f(x)$

Substitute $x = 6$ into $\frac{5x+1}{x + 25}$. We get $\frac{5\times6+1}{6 + 25}=\frac{30 + 1}{31}=\frac{31}{31}=1$.

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

Substitute $x = 0$ into $\frac{5x+1}{x + 25}$. We get $\frac{5\times0+1}{0 + 25}=\frac{1}{25}$.

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

Substitute $x = 1$ into $\frac{5x+1}{x + 25}$. We get $\frac{5\times1+1}{1 + 25}=\frac{6}{26}=\frac{3}{13}$.

Answer:

(A) A. $\lim_{x\rightarrow6}\frac{5x^{2}-29x - 6}{x^{2}+19x - 150}=1$ (B) A. $\lim_{x\rightarrow0}\frac{5x^{2}-29x - 6}{x^{2}+19x - 150}=\frac{1}{25}$ (C) A. $\lim_{x\rightarrow1}\frac{5x^{2}-29x - 6}{x^{2}+19x - 150}=\frac{3}{13}$