let f(x) = (x - 3)/(x² - 3x). find the indicated quantities, if they exist. (a) lim f(x) as x→0 (b) lim f(x)…

let f(x) = (x - 3)/(x² - 3x). find the indicated quantities, if they exist. (a) lim f(x) as x→0 (b) lim f(x) as x→3 (c) lim f(x) as x→9 (a) select the correct choice below and, if necessary, fill in the answer box to complete your choice. o a. lim (x - 3)/(x² - 3x) as x→0 = (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 - 3)/(x² - 3x) as x→3 = (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 - 3)/(x² - 3x) as x→9 = (type an integer or a simplified fraction.) o b. the limit does not exist.
Answer
Explanation:
Step1: Simplify the function
First, factor the denominator of $f(x)=\frac{x - 3}{x^{2}-3x}=\frac{x - 3}{x(x - 3)}$. For $x\neq3$, we can cancel out the common - factor $(x - 3)$ and get $f(x)=\frac{1}{x}$ ($x\neq3$).
Step2: Calculate $\lim_{x\rightarrow0}f(x)$
Substitute $x = 0$ into $\frac{1}{x}$. As $x\rightarrow0$, $\frac{1}{x}$ approaches positive or negative infinity depending on the direction of approach. So, $\lim_{x\rightarrow0}\frac{x - 3}{x^{2}-3x}$ does not exist.
Step3: Calculate $\lim_{x\rightarrow3}f(x)$
Since $f(x)=\frac{1}{x}$ for $x\neq3$, we substitute $x = 3$ into $\frac{1}{x}$. Then $\lim_{x\rightarrow3}\frac{x - 3}{x^{2}-3x}=\lim_{x\rightarrow3}\frac{1}{x}=\frac{1}{3}$.
Step4: Calculate $\lim_{x\rightarrow9}f(x)$
Substitute $x = 9$ into $\frac{1}{x}$. Then $\lim_{x\rightarrow9}\frac{x - 3}{x^{2}-3x}=\lim_{x\rightarrow9}\frac{1}{x}=\frac{1}{9}$.
Answer:
(A) B. The limit does not exist. (B) A. $\frac{1}{3}$ (C) A. $\frac{1}{9}$