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) = square ) (simplify your answer.)

Explanation:

Step1: Factor the numerator and denominator

First, factor the numerator \(x^2 - 9x + 18\) and the denominator \(x^2 - 3x\).
The numerator factors as \((x - 3)(x - 6)\) (since \( -3 \times -6 = 18\) and \( -3 + (-6) = -9\)).
The denominator factors as \(x(x - 3)\) (by factoring out an \(x\)).
So, \(f(x)=\frac{(x - 3)(x - 6)}{x(x - 3)}\), and we can cancel out the \((x - 3)\) terms (for \(x
eq3\)) to get \(f(x)=\frac{x - 6}{x}\) (with the restriction \(x
eq0\) and \(x
eq3\)).

Step2: Find the limit as \(x\to3^-\)

Now, we want to find \(\lim_{x\to3^-} f(x)\). Since we have simplified \(f(x)\) to \(\frac{x - 6}{x}\) for \(x
eq3\), we can directly substitute \(x = 3\) into this simplified function (because the limit as \(x\) approaches 3 does not depend on the value of the function at \(x = 3\), only the values around \(x = 3\)).
Substitute \(x = 3\) into \(\frac{x - 6}{x}\):
\(\frac{3 - 6}{3}=\frac{-3}{3}=-1\)

Answer:

\(-1\)