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) = square ) (simplify your answer.)

Explanation:

Step1: Factor numerator and denominator

Factor \(x^2 - 9x + 18=(x - 3)(x - 6)\) and \(x^2 - 3x=x(x - 3)\). So \(f(x)=\frac{(x - 3)(x - 6)}{x(x - 3)}\) (for \(x eq3\)). Cancel \(x - 3\) ( \(x eq3\) ), get \(f(x)=\frac{x - 6}{x}\) ( \(x eq0,3\) ).

Step2: Find \(\lim_{x

ightarrow0^-}f(x)\)
Substitute \(f(x)=\frac{x - 6}{x}\) into the limit. As \(x ightarrow0^-\), \(x\) is negative and approaches 0, \(x - 6 ightarrow - 6\). So \(\lim_{x ightarrow0^-}\frac{x - 6}{x}=\lim_{x ightarrow0^-}(1-\frac{6}{x})\). As \(x ightarrow0^-\), \(\frac{6}{x} ightarrow-\infty\), so \(1-\frac{6}{x} ightarrow+\infty\)? Wait, no, wait: \(\frac{x - 6}{x}=\frac{x}{x}-\frac{6}{x}=1-\frac{6}{x}\). When \(x ightarrow0^-\), \(\frac{6}{x}\) is negative large (since \(x\) is negative), so \(-\frac{6}{x}\) is positive large. Wait, no, let's do it directly: \(\lim_{x ightarrow0^-}\frac{x - 6}{x}=\lim_{x ightarrow0^-}(1-\frac{6}{x})\). As \(x ightarrow0^-\), \(\frac{6}{x} ightarrow-\infty\), so \(-\frac{6}{x} ightarrow+\infty\), so \(1+\infty=+\infty\)? Wait, no, wait, original function after canceling: \(f(x)=\frac{x - 6}{x}\), domain \(x eq0,3\). So as \(x ightarrow0^-\), \(x\) is a small negative number, numerator \(x - 6 ightarrow - 6\) (negative), denominator \(x\) is negative. So negative divided by negative is positive. And as \(x ightarrow0^-\), \(|x|\) is small, so \(\frac{|x - 6|}{|x|}\approx\frac{6}{|x|}\), so the limit is \(+\infty\)? Wait, but maybe I made a mistake. Wait, let's check the factoring again. \(x^2-9x + 18=(x - 3)(x - 6)\), correct. \(x^2-3x=x(x - 3)\), correct. So cancel \(x - 3\) ( \(x eq3\) ), so \(f(x)=\frac{x - 6}{x}\), \(x eq0,3\). So \(\lim_{x ightarrow0^-}\frac{x - 6}{x}\). Let's take \(x=-0.1\), then \(\frac{-0.1 - 6}{-0.1}=\frac{-6.1}{-0.1}=61\). Take \(x=-0.01\), \(\frac{-0.01 - 6}{-0.01}=\frac{-6.01}{-0.01}=601\). So as \(x ightarrow0^-\), the limit is \(+\infty\)? Wait, but the problem says "simplify your answer". Wait, maybe I misread the limit. Wait, the first limit is \(\lim_{x ightarrow0^-}f(x)\). Wait, but let's check again. Wait, maybe the function is \(\frac{x^2 - 9x + 18}{x^2 - 3x}\), so when \(x ightarrow0^-\), denominator \(x^2 - 3x=x(x - 3)\), as \(x ightarrow0^-\), \(x\) is negative, \(x - 3\) is negative, so denominator is positive (negative times negative). Numerator \(x^2 - 9x + 18\), as \(x ightarrow0\), numerator is \(0 - 0 + 18 = 18\) (positive). Wait, wait, I made a mistake in factoring the numerator? Wait, \(x^2-9x + 18\): discriminant \(81 - 72 = 9\), roots \(\frac{9\pm3}{2}\), so 6 and 3. So \((x - 3)(x - 6)\), correct. So numerator at \(x ightarrow0\) is \((-3)(-6)=18\) (positive). Denominator \(x(x - 3)\), as \(x ightarrow0^-\), \(x\) is negative, \(x - 3\) is negative, so denominator is positive (negative * negative = positive). Wait, so numerator is positive, denominator is positive, so the limit should be \(\frac{18}{0^+}\)? Wait, no, when \(x ightarrow0^-\), \(x\) is negative, so \(x(x - 3)=x^2 - 3x\), when \(x=-0.1\), denominator is \((-0.1)^2 - 3*(-0.1)=0.01 + 0.3 = 0.31\) (positive). Numerator is \((-0.1)^2 - 9*(-0.1)+18=0.01 + 0.9 + 18 = 18.91\) (positive). So \(\frac{18.91}{0.31}\approx61\), as \(x ightarrow0^-\), denominator approaches 0 from the positive side (since \(x(x - 3)\) when \(x ightarrow0^-\) is \(x\) negative, \(x - 3\) negative, product positive, approaching 0). Numerator approaches 18 (positive). So the limit is \(+\infty\)? But the problem says "simplify your answer". Wait, maybe I messed up the sign of the denominator. Wait, \(x^2 - 3x=x(x - 3)\). When \(x ightarrow0^-\), \(x\) is negative, \(x - 3\) is negative (since \(x<0<3\)), so \(x(x - 3)\) is positive (negative times negative). So denominator approaches 0 from the positive side, numerator approaches 18 (positive), so the limit is \(+\infty\). But maybe the problem expects a different approach. Wait, no, let's re-express \(f(x)=\frac{(x - 3)(x - 6)}{x(x - 3)}=\frac{x - 6}{x}\) for \(x eq3\). So \(\lim_{x ightarrow0^-}\frac{x - 6}{x}=\lim_{x ightarrow0^-}(1-\frac{6}{x})\). As \(x ightarrow0^-\), \(\frac{6}{x}\) is negative (since \(x\) is negative), so \(-\frac{6}{x}\) is positive, and as \(x ightarrow0^-\), \(\frac{6}{x} ightarrow-\infty\), so \(-\frac{6}{x} ightarrow+\infty\), so \(1+\infty=+\infty\). So the limit is \(+\infty\).

Answer:

\(\lim_{x
ightarrow0^-}f(x)=\infty\) (or more precisely \(+\infty\))