Question

exercises 4.6 limits at infinity and asymptotes
score: 9/15 answered: 9/15
question 10
evaluate the following limits. if needed, enter oo for \\(\infty\\) and -oo for \\(-\infty\\).
(a) \\(\lim_{x \to \frac{3}{6}^+} \frac{-24x}{3 - 6x} = \\)
(b) \\(\lim_{x \to \frac{3}{6}^-} \frac{-24x}{3 - 6x} = \\)

Explanation:

Step1: Analyze the denominator at \( x = \frac{3}{6}=\frac{1}{2} \)

First, simplify the denominator \( 3 - 6x \). When \( x=\frac{1}{2} \), \( 3-6\times\frac{1}{2}=3 - 3 = 0 \). So we need to analyze the sign of the denominator as \( x \) approaches \( \frac{1}{2} \) from the right (\( x ightarrow\frac{1}{2}^+ \)) and left (\( x ightarrow\frac{1}{2}^- \)).

Step2: Analyze part (a): \( x

ightarrow\frac{1}{2}^+ \)
For \( x ightarrow\frac{1}{2}^+ \), \( x>\frac{1}{2} \), so \( 6x>3 \), then \( 3 - 6x<0 \) (since \( 6x - 3>0 \), so \( 3 - 6x=-(6x - 3)<0 \)). The numerator is \( - 24x \), when \( x ightarrow\frac{1}{2}^+ \), numerator \( -24x ightarrow - 24\times\frac{1}{2}=-12 \) (a negative number, but let's focus on the sign and the behavior). Now, let's rewrite the function: \( \frac{-24x}{3 - 6x}=\frac{-24x}{- (6x - 3)}=\frac{24x}{6x - 3}=\frac{24x}{6(x-\frac{1}{2})}=\frac{4x}{x - \frac{1}{2}} \). As \( x ightarrow\frac{1}{2}^+ \), \( x-\frac{1}{2} ightarrow0^+ \) (positive and small), and \( 4x ightarrow4\times\frac{1}{2} = 2 \) (positive). So a positive number divided by a positive small number approaches \( +\infty \)? Wait, wait, let's check again. Wait, original numerator: \( -24x \), denominator: \( 3 - 6x \). When \( x ightarrow\frac{1}{2}^+ \), denominator \( 3 - 6x=3-6x \), if \( x=\frac{1}{2}+h \), \( h ightarrow0^+ \), then denominator \( 3-6(\frac{1}{2}+h)=3 - 3 - 6h=-6h ightarrow0^- \) (negative small). Numerator: \( -24(\frac{1}{2}+h)=-12-24h ightarrow - 12 \) (negative). So negative divided by negative: \( \frac{-12 - 24h}{-6h}=\frac{12 + 24h}{6h}=\frac{2 + 4h}{h}=\frac{2}{h}+4 \). As \( h ightarrow0^+ \), \( \frac{2}{h} ightarrow+\infty \), so the limit is \( +\infty \)? Wait, no, wait, let's do it simpler. Let's take the limit as \( x ightarrow\frac{1}{2} \), factor numerator and denominator. The function is \( \frac{-24x}{3 - 6x}=\frac{-24x}{-6(x - \frac{1}{2})}=\frac{4x}{x - \frac{1}{2}} \). Now, as \( x ightarrow\frac{1}{2}^+ \), \( x-\frac{1}{2} ightarrow0^+ \), so \( \frac{4x}{x - \frac{1}{2}} ightarrow\frac{4\times\frac{1}{2}}{0^+}=\frac{2}{0^+} ightarrow+\infty \). Wait, but let's check with values. Let \( x = 0.6 \) (which is \( \frac{3}{5}>\frac{1}{2}=0.5 \)). Then denominator: \( 3-6\times0.6=3 - 3.6=-0.6 \), numerator: \( -24\times0.6=-14.4 \). Then \( \frac{-14.4}{-0.6}=24 \). If \( x = 0.51 \), denominator: \( 3-6\times0.51=3 - 3.06=-0.06 \), numerator: \( -24\times0.51=-12.24 \), \( \frac{-12.24}{-0.06}=204 \). As \( x \) approaches 0.5 from the right, the value increases, so it approaches \( +\infty \)? Wait, no, wait my first rewrite was wrong. Wait \( 3 - 6x=-6x + 3=-6(x-\frac{1}{2}) \). So numerator: \( -24x \), denominator: \( -6(x - \frac{1}{2}) \). So \( \frac{-24x}{-6(x - \frac{1}{2})}=\frac{4x}{x - \frac{1}{2}} \). Now, when \( x ightarrow\frac{1}{2}^+ \), \( x-\frac{1}{2}\) is positive and small, \( 4x \) is positive (since \( x>0 \)), so positive divided by positive small: \( \frac{4x}{x - \frac{1}{2}}\approx\frac{4\times\frac{1}{2}}{\epsilon}=\frac{2}{\epsilon}\), where \( \epsilon ightarrow0^+ \), so this approaches \( +\infty \). Wait, but when I plugged \( x = 0.6 \), I got 24, \( x = 0.51 \), got 204, which is increasing, so yes, approaches \( +\infty \). Wait, but let's check the sign again. Original numerator: \( -24x \), denominator: \( 3 - 6x \). When \( x>\frac{1}{2} \), denominator is negative (3 - 6x < 0), numerator is negative (since x positive, -24x negative). Negative divided by negative is positive. And as x approaches 0.5 from the right, denominator approaches 0 from the negative side, numerator approaches -12 (negative). So negative / negative = positive, and the magnitude: |numerator| approaches 12, |denominator| approaches 0, so the whole thing approaches \( +\infty \).

Step3: Analyze part (b): \( x

ightarrow\frac{1}{2}^- \)
For \( x ightarrow\frac{1}{2}^- \), \( x<\frac{1}{2} \), so \( 6x<3 \), then \( 3 - 6x>0 \) (since \( 3 - 6x=3-6x \), if \( x=\frac{1}{2}-h \), \( h ightarrow0^+ \), then denominator \( 3-6(\frac{1}{2}-h)=3 - 3 + 6h=6h ightarrow0^+ \) (positive small). Numerator: \( -24x=-24(\frac{1}{2}-h)=-12 + 24h ightarrow - 12 \) (negative). So negative divided by positive: \( \frac{-12 + 24h}{6h}=\frac{-12}{6h}+\frac{24h}{6h}=\frac{-2}{h}+4 \). As \( h ightarrow0^+ \), \( \frac{-2}{h} ightarrow - \infty \), so the limit is \( -\infty \)? Wait, let's use the rewritten function: \( \frac{4x}{x - \frac{1}{2}} \). When \( x ightarrow\frac{1}{2}^- \), \( x-\frac{1}{2} ightarrow0^- \) (negative small), \( 4x ightarrow2 \) (positive). So positive divided by negative small: \( \frac{2}{\epsilon} \) where \( \epsilon ightarrow0^- \), so this approaches \( -\infty \). Let's check with values. Let \( x = 0.4 \) (which is \( <0.5 \)), denominator: \( 3-6\times0.4=3 - 2.4 = 0.6 \) (positive), numerator: \( -24\times0.4=-9.6 \). Then \( \frac{-9.6}{0.6}=-16 \). If \( x = 0.49 \), denominator: \( 3-6\times0.49=3 - 2.94 = 0.06 \) (positive), numerator: \( -24\times0.49=-11.76 \). Then \( \frac{-11.76}{0.06}=-196 \). As \( x \) approaches 0.5 from the left, the value decreases (more negative), so approaches \( -\infty \).

Wait, but let's re-express the original function:

\( \lim_{x ightarrow\frac{1}{2}^+}\frac{-24x}{3 - 6x} \)

Factor denominator: \( 3 - 6x = -6(x - \frac{1}{2}) \)

So the function becomes \( \frac{-24x}{-6(x - \frac{1}{2})}=\frac{4x}{x - \frac{1}{2}} \)

Now, as \( x ightarrow\frac{1}{2}^+ \), \( x - \frac{1}{2} ightarrow0^+ \), so \( \frac{4x}{x - \frac{1}{2}}\approx\frac{4\times\frac{1}{2}}{\epsilon}=\frac{2}{\epsilon} ightarrow+\infty \) (since \( \epsilon>0 \) small)

As \( x ightarrow\frac{1}{2}^- \), \( x - \frac{1}{2} ightarrow0^- \), so \( \frac{4x}{x - \frac{1}{2}}\approx\frac{2}{\epsilon} ightarrow-\infty \) (since \( \epsilon<0 \) small)

Wait, but let's check the numerator and denominator signs again for part (a):

Numerator: \( -24x \), when \( x ightarrow\frac{1}{2}^+ \), \( x \) is positive, so numerator is negative.

Denominator: \( 3 - 6x \), when \( x ightarrow\frac{1}{2}^+ \), \( x>\frac{1}{2} \), so \( 6x>3 \), so \( 3 - 6x<0 \) (negative).

So negative (numerator) divided by negative (denominator) is positive. And as denominator approaches 0 from negative side, numerator approaches -12 (negative). So |numerator| approaches 12, |denominator| approaches 0, so the fraction approaches \( +\infty \) (since positive and the magnitude increases).

For part (b):

Numerator: negative (approaches -12), denominator: positive (approaches 0 from positive side). So negative divided by positive is negative, and magnitude increases, so approaches \( -\infty \).

Wait, maybe a better way: divide numerator and denominator by x (for limits at infinity, but here it's a finite limit, so we can analyze the sign and the behavior as denominator approaches 0).

Alternative approach for (a):

\( \lim_{x ightarrow\frac{1}{2}^+}\frac{-24x}{3 - 6x} \)

Let \( t=x - \frac{1}{2} \), so \( x=\frac{1}{2}+t \), \( t ightarrow0^+ \)

Substitute:

\( \lim_{t ightarrow0^+}\frac{-24(\frac{1}{2}+t)}{3 - 6(\frac{1}{2}+t)}=\lim_{t ightarrow0^+}\frac{-12 - 24t}{3 - 3 - 6t}=\lim_{t ightarrow0^+}\frac{-12 - 24t}{-6t}=\lim_{t ightarrow0^+}\frac{12 + 24t}{6t}=\lim_{t ightarrow0^+}(\frac{12}{6t}+\frac{24t}{6t})=\lim_{t ightarrow0^+}(\frac{2}{t}+4) \)

As \( t ightarrow0^+ \), \( \frac{2}{t} ightarrow+\infty \), so the limit is \( +\infty \)

For (b):

\( t=x - \frac{1}{2} \), \( t ightarrow0^- \)

Substitute:

\( \lim_{t ightarrow0^-}\frac{-24(\frac{1}{2}+t)}{3 - 6(\frac{1}{2}+t)}=\lim_{t ightarrow0^-}\frac{-12 - 24t}{-6t}=\lim_{t ightarrow0^-}(\frac{2}{t}+4) \)

As \( t ightarrow0^- \), \( \frac{2}{t} ightarrow-\infty \), so the limit is \( -\infty \)

Answer:

(a) \( \infty \) (or \( +\infty \))
(b) \( -\infty \)