Question

let ( f(x) = \frac{x^2 - 9x + 18}{x^2 - 3x} ). complete parts (a) through (c) below.

a. analyze ( limlimits_{x \to 0^-} f(x) ), ( limlimits_{x \to 0^+} f(x) ), ( limlimits_{x \to 3^-} f(x) ), and ( limlimits_{x \to 3^+} f(x) ).

( limlimits_{x \to 0^-} f(x) = infty ) (simplify your answer.)

( limlimits_{x \to 0^+} f(x) = -infty ) (simplify your answer.)

( limlimits_{x \to 3^-} f(x) = -1 ) (simplify your answer.)

( limlimits_{x \to 3^+} f(x) = square ) (simplify your answer.)

Explanation:

Step1: Simplify the function

First, factor the numerator and the denominator.
The numerator \(x^2 - 9x + 18\) can be factored as \((x - 3)(x - 6)\).
The denominator \(x^2 - 3x\) can be factored as \(x(x - 3)\).
So, \(f(x)=\frac{(x - 3)(x - 6)}{x(x - 3)}\), for \(x
eq3\), we can cancel out the \((x - 3)\) terms, getting \(f(x)=\frac{x - 6}{x}\) (where \(x
eq0\) and \(x
eq3\)).

Step2: Find the limit as \(x

ightarrow3^{+}\)
Now, we find \(\lim_{x
ightarrow3^{+}}f(x)\). Substitute \(x = 3\) into the simplified function \(\frac{x - 6}{x}\) (since the limit as \(x
ightarrow3^{+}\) does not depend on the value at \(x = 3\), and the simplified function is valid for \(x\) near 3 except at \(x = 3\) itself).
\(\lim_{x
ightarrow3^{+}}\frac{x - 6}{x}=\frac{3 - 6}{3}=\frac{-3}{3}=- 1\)

Answer:

\(-1\)