Question

let (f(x)=\frac{x - 6}{x^{2}-6x}). find the indicated quantities, if they exist.
(a) (lim_{x
ightarrow0}f(x)) (b) (lim_{x
ightarrow6}f(x)) (c) (lim_{x
ightarrow36}f(x)
(a) select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a. (lim_{x
ightarrow0}\frac{x - 6}{x^{2}-6x}=square) (type an integer or a simplified fraction.)
b. the limit does not exist

Explanation:

Step1: Simplify the function

First, factor the denominator of \(f(x)=\frac{x - 6}{x^{2}-6x}\). Since \(x^{2}-6x=x(x - 6)\), then \(f(x)=\frac{x - 6}{x(x - 6)}=\frac{1}{x}\), for \(x
eq6\).

Step2: Calculate \(\lim_{x

ightarrow0}f(x)\)
Substitute \(x = 0\) into \(\frac{1}{x}\). As \(x
ightarrow0\), \(\frac{1}{x}\) approaches either positive or negative infinity depending on the side - approach. So \(\lim_{x
ightarrow0}\frac{x - 6}{x^{2}-6x}\) does not exist.

Step3: Calculate \(\lim_{x

ightarrow6}f(x)\)
Substitute \(x = 6\) into \(\frac{1}{x}\). \(\lim_{x
ightarrow6}\frac{1}{x}=\frac{1}{6}\).

Step4: Calculate \(\lim_{x

ightarrow36}f(x)\)
Substitute \(x = 36\) into \(\frac{1}{x}\). \(\lim_{x
ightarrow36}\frac{1}{x}=\frac{1}{36}\).

Answer:

(A) B. The limit does not exist
(B) \(\frac{1}{6}\)
(C) \(\frac{1}{36}\)