let f(x) = (x - 5)/(x² - 5x). find the indicated quantities, if they exist. (a) lim f(x) as x→0 (b) lim f(x)…

let f(x) = (x - 5)/(x² - 5x). find the indicated quantities, if they exist. (a) lim f(x) as x→0 (b) lim f(x) as x→5 (c) lim f(x) as x→25 (a) select the correct choice below and, if necessary, fill in the answer box to complete your choice. a. lim (x - 5)/(x² - 5x) as x→0 = (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 - 5)/(x² - 5x) as x→5 = (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 - 5)/(x² - 5x) as x→25 = (type an integer or a simplified fraction.) b. the limit does not exist.

let f(x) = (x - 5)/(x² - 5x). find the indicated quantities, if they exist. (a) lim f(x) as x→0 (b) lim f(x) as x→5 (c) lim f(x) as x→25 (a) select the correct choice below and, if necessary, fill in the answer box to complete your choice. a. lim (x - 5)/(x² - 5x) as x→0 = (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 - 5)/(x² - 5x) as x→5 = (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 - 5)/(x² - 5x) as x→25 = (type an integer or a simplified fraction.) b. the limit does not exist.

Answer

Explanation:

Step1: Simplify the function

First, factor the denominator of $f(x)=\frac{x - 5}{x^{2}-5x}=\frac{x - 5}{x(x - 5)}$. For $x\neq5$, we can cancel out the common - factor $(x - 5)$ and get $f(x)=\frac{1}{x}$ ($x\neq5$).

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 $x$ approaching 0. So, $\lim_{x\rightarrow0}\frac{x - 5}{x^{2}-5x}$ does not exist.

Step3: Calculate $\lim_{x\rightarrow5}f(x)$

Since $f(x)=\frac{1}{x}$ for $x\neq5$, then $\lim_{x\rightarrow5}\frac{x - 5}{x^{2}-5x}=\lim_{x\rightarrow5}\frac{1}{x}=\frac{1}{5}$.

Step4: Calculate $\lim_{x\rightarrow25}f(x)$

Since $f(x)=\frac{1}{x}$ for $x\neq5$, then $\lim_{x\rightarrow25}\frac{x - 5}{x^{2}-5x}=\lim_{x\rightarrow25}\frac{1}{x}=\frac{1}{25}$.

Answer:

(A) B. The limit does not exist. (B) A. $\lim_{x\rightarrow5}\frac{x - 5}{x^{2}-5x}=\frac{1}{5}$ (C) A. $\lim_{x\rightarrow25}\frac{x - 5}{x^{2}-5x}=\frac{1}{25}$