Consider the parameterized rational inequality $\frac{x - 2}{\epsilon x + 1} > 0$ where $\epsilon > 0$ is a small positive parameter. Determine the solution set in interval notation in terms of $\epsilon$, and describe how the boundary points and shaded intervals on the real number line behave asymptotically in the extreme limit as $\epsilon \to 0^+$.

Consider the parameterized rational inequality $\frac{x - 2}{\epsilon x + 1} > 0$ where $\epsilon > 0$ is a small positive parameter. Determine the solution set in interval notation in terms of $\epsilon$ , and describe how the boundary points and shaded intervals on the real number line behave asymptotically in the extreme limit as $\epsilon \to 0^+$ .

Final answer

For $\epsilon > 0$ , the solution set is $(-\infty, -\frac{1}{\epsilon}) \cup (2, \infty)$ . As $\epsilon \to 0^+$ , the boundary $-\frac{1}{\epsilon} \to -\infty$ , causing the left branch to vanish asymptotically and leaving the limiting solution set $(2, \infty)$ , represented on the number line by an open circle at $x = 2$ with shading extending indefinitely to the right.

Steps

Identify the Critical Values

To solve the rational inequality $\frac{x - 2}{\epsilon x + 1} > 0$ find the critical points where the numerator or denominator equals zero:

Numerator: $x - 2 = 0 \implies x = 2$
Denominator: $\epsilon x + 1 = 0 \implies x = -\frac{1}{\epsilon}$

Because $\epsilon > 0$ , the quantity $-\frac{1}{\epsilon}$ is strictly negative. Comparing the two boundary values: $-\frac{1}{\epsilon} < 0 < 2$

Determine the Sign in Each Interval

The two critical numbers divide the real number line into three intervals: $(-\infty, -\frac{1}{\epsilon}), \quad (-\frac{1}{\epsilon}, 2), \quad (2, \infty)$

Test the sign of the quotient $\frac{x - 2}{\epsilon x + 1}$ across these intervals:

For $x \in (-\infty, -\frac{1}{\epsilon})$ : Both $x - 2 < 0$ and $\epsilon x + 1 < 0$ . A negative divided by a negative is positive: $\frac{(-)}{(-)} > 0$
For $x \in (-\frac{1}{\epsilon}, 2)$ : Here, $x - 2 < 0$ while $\epsilon x + 1 > 0$ . A negative divided by a positive is negative: $\frac{(-)}{(+)} < 0$
For $x \in (2, \infty)$ : Both $x - 2 > 0$ and $\epsilon x + 1 > 0$ . A positive divided by a positive is positive: $\frac{(+)}{(+)} > 0$

The inequality is strictly greater than zero ( $> 0$ ), so neither boundary point is included. Thus, for any fixed $\epsilon > 0$ , the solution set in interval notation is: $\left(-\infty, -\frac{1}{\epsilon}\right) \cup (2, \infty)$

Analyze the Asymptotic Limit as $\epsilon \to 0^+$

Consider the behavior of the boundaries on the real number line as $\epsilon$ approaches $0$ through positive values:

Right Boundary ( $x = 2$ ): This boundary is constant and independent of $\epsilon$ . It remains fixed at $x = 2$ as an open circle, with shading extending indefinitely to the right, $(2, \infty)$ .
Left Boundary ( $x = -\frac{1}{\epsilon}$ ): As $\epsilon \to 0^+$ , the denominator approaches $0$ through positive numbers, so: $-\frac{1}{\epsilon} \to -\infty$

As $-\frac{1}{\epsilon}$ moves infinitely far to the left, the interval $(-\infty, -\frac{1}{\epsilon})$ is pushed entirely off any finite window of the real line, effectively vanishing.

Therefore, in the extreme limit as $\epsilon \to 0^+$ , the inequality reduces to the linear inequality $x - 2 > 0$ , leaving the limiting solution set: $(2, \infty)$ On the number line, this is represented by an open circle at $x = 2$ with shading extending indefinitely toward $+\infty$ .

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student clears the denominator by multiplying by εx + 1 without considering that it can be negative, losing the left interval (-∞, -1/ε).

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
Multiply both sides by $\epsilon$ x + 1: $\implies$ x - 2 > 0 $\implies$ x > 2
Correct
$\text{Find critical points where numerator or denominator equals }$ 0: x - 2 = 0 $\implies$ x = 2, $\epsilon$ x + 1 = 0 $\implies$ x = - $\frac{1}{\epsilon}$
You cannot multiply both sides of an inequality by an expression containing a variable without accounting for its sign. Because $\epsilon x + 1$ is negative when $x < -\frac{1}{\epsilon}$ , multiplying by it would reverse the inequality sign. Instead, find the critical points of both the numerator and denominator to partition the number line into test intervals.
Step 2
Solution set: (2, $\infty$ )
Correct
$\left(-\infty, -\frac{1}{\epsilon}\right) \cup (2, \infty)$
Because the critical point at $x = -\frac{1}{\epsilon}$ was excluded in the first step, this solution set misses the region $(-\infty, -\frac{1}{\epsilon})$ , where both the numerator and denominator are negative, making their quotient positive.
Step 3
Boundary point is x = 2, which does not depend on $\epsilon$ . $\implies$ As $\epsilon$ $\to$ 0^+, the boundary point x = 2 remains fixed and the shaded interval (2, $\infty$ ) remains unchanged.
Correct
$\text{Boundary }$ x = 2 $\text{ is fixed. As }$ $\epsilon$ $\to$ 0^+, - $\frac{1}{\epsilon}$ $\to$ - $\infty$ , $\text{ so the interval }$ $\left(-\infty, -\frac{1}{\epsilon}\right)$ $\text{ moves infinitely far left and vanishes asymptotically, leaving }$ (2, $\infty$ ).
You correctly observed that the boundary $x = 2$ does not depend on $\epsilon$ . However, because the boundary $-\frac{1}{\epsilon}$ was missing from your solution set, the asymptotic analysis did not capture how the left interval shifts infinitely far to the left and vanishes as $\epsilon \to 0^+$ .

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly identified that $x = 2$ makes the numerator zero.

You accurately noted that the boundary point $x = 2$ is independent of $\epsilon$ .

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Clearing variable denominator in inequality

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Treating a rational inequality like a linear equation by multiplying across by a variable denominator without accounting for the fact that the denominator changes sign across its own zero.

REASONING PATH

Your ACEpath teacher’s suggested approach

Variable Denominator Rule: Never multiply or divide an inequality by an algebraic expression unless you know its sign with certainty.
Set both numerator and denominator equal to zero to identify all critical boundary points on the real line.
Sign Chart Method: A rational expression $\frac{P(x)}{Q(x)}$ changes sign only at the roots of $P(x)$ and $Q(x)$ .
Plot all critical points on a number line and test a sample point in each interval to find where the quotient is strictly positive.

CORRECTED PATH

ACEpath Teacher's Demonstration

Set numerator and denominator to zero: $x - 2 = 0 \implies x = 2$ and $\epsilon x + 1 = 0 \implies x = -\frac{1}{\epsilon}$ .
Since $\epsilon > 0$ , the critical points satisfy $-\frac{1}{\epsilon} < 2$ , partitioning the real line into $\left(-\infty, -\frac{1}{\epsilon}\right)$ , $\left(-\frac{1}{\epsilon}, 2\right)$ , and $(2, \infty)$ .
Test the sign in each interval: for $x < -\frac{1}{\epsilon}$ , $\frac{(-)}{(-)} > 0$ ; for $-\frac{1}{\epsilon} < x < 2$ , $\frac{(-)}{(+)} < 0$ ; for $x > 2$ , $\frac{(+)}{(+)} > 0$ .
Combine the positive intervals: the solution set is $\left(-\infty, -\frac{1}{\epsilon}\right) \cup (2, \infty)$ .
Evaluate the limit $\epsilon \to 0^+$ : $-\frac{1}{\epsilon} \to -\infty$ , pushing the interval $\left(-\infty, -\frac{1}{\epsilon}\right)$ off to $-\infty$ and leaving $(2, \infty)$ on any finite portion of the number line.

Final answer

For $\epsilon > 0$ , the solution set is $\left(-\infty, -\frac{1}{\epsilon}\right) \cup (2, \infty)$ . As $\epsilon \to 0^+$ , the boundary $-\frac{1}{\epsilon} \to -\infty$ , so the left interval vanishes asymptotically to $-\infty$ , leaving the limiting solution set $(2, \infty)$ with an open circle at $x = 2$ and shading extending indefinitely to the right.

NEXT STEPS

Final Takeaways from Your Teacher

Practice solving rational inequalities by finding zeros of both numerator and denominator rather than clearing fractions.
Draw sign charts or number-line test intervals before writing down solution sets.
Review parameter limits where critical boundaries migrate toward $\pm\infty$ .

OverviewACEpath Teacher's Diagnosis

02 The student mistakenly solves εx + 1 = 0 as x = -ε instead of x = -1/ε, leading to an incorrect boundary point that vanishes rather than diverging as ε → 0⁺.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
Numerator gives $x = 2$
Step 2
Denominator gives $\epsilon x + 1 = 0 \implies x = -\epsilon$
Correct
$\epsilon x + 1 = 0 \implies \epsilon x = -1 \implies x = -\frac{1}{\epsilon}$
When solving $\epsilon x + 1 = 0$ , subtracting $1$ gives $\epsilon x = -1$ . You need to divide both sides by the coefficient $\epsilon$ , which yields $x = -\frac{1}{\epsilon}$ , not $x = -\epsilon$ .
Step 3
Since $\epsilon > 0$ , $-\epsilon < 2 \implies \text{Solution set: } (-\infty, -\epsilon) \cup (2, \infty)$
Correct
Since $\epsilon > 0$ , $-\frac{1}{\epsilon} < 0 < 2 \implies \text{Solution set: } \left(-\infty, -\frac{1}{\epsilon}\right) \cup (2, \infty)$
Your inequality sign testing and interval union logic followed correctly from your boundary points, but the left interval inherited the incorrect boundary point.
Step 4
As $\epsilon \to 0^+$ , the boundary point $x = -\epsilon$ approaches $0$ , while $x = 2$ remains fixed. The left shaded interval $(-\infty, -\epsilon)$ expands rightward toward $(-\infty, 0)$ , and the interval $(2, \infty)$ stays fixed.
Correct
As $\epsilon \to 0^+$ , the boundary point $-\frac{1}{\epsilon} \to -\infty$ . The left interval $\left(-\infty, -\frac{1}{\epsilon}\right)$ gets pushed infinitely far left and vanishes asymptotically, leaving only the right interval $(2, \infty)$ .
Because of the boundary error, you analyzed the limit of $-\epsilon$ as $\epsilon \to 0^+$ , whereas the true boundary $-\frac{1}{\epsilon}$ diverges toward $-\infty$ .

WHAT YOU DID WELL

Compliments from Your Teacher

You identified that the critical values come from setting the numerator and denominator to zero.

You correctly found the numerator root at $x = 2$ .

You maintained logical consistency by properly structuring the solution intervals based on your calculated boundary points.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Algebraic isolation error with parameterized coefficient

You solved $\epsilon x + 1 = 0$ as $x = -\epsilon$ instead of dividing by $\epsilon$ to get $x = -\frac{1}{\epsilon}$ .

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Treating the multiplication $\epsilon x$ as if the parameter were an additive term or confusing multiplication by $\epsilon$ with division by $\epsilon$ when isolating $x$ .

REASONING PATH

Your ACEpath teacher’s suggested approach

Division property of equality: To isolate $x$ in $ax + b = 0$ , subtract $b$ and then divide by the coefficient $a$ : $x = -\frac{b}{a}$ .
Always explicitly write out the step $\epsilon x = -1$ before dividing both sides by $\epsilon$ .
Reciprocal limit behavior: For a small positive parameter $\epsilon \to 0^+$ , $\frac{1}{\epsilon} \to +\infty$ , so $-\frac{1}{\epsilon} \to -\infty$ .
Check whether boundary points diverge or vanish as parameters become arbitrarily small.

CORRECTED PATH

ACEpath Teacher's Demonstration

Subtract $1$ from both sides: $\epsilon x = -1$ .
Divide by $\epsilon$ (where $\epsilon > 0$ ): $x = -\frac{1}{\epsilon}$ .
Because $\epsilon > 0$ , $-\frac{1}{\epsilon} < 0 < 2$ . Testing intervals gives the solution set $\left(-\infty, -\frac{1}{\epsilon}\right) \cup (2, \infty)$ .
Evaluate the limit as $\epsilon \to 0^+$ : $-\frac{1}{\epsilon} \to -\infty$ . The left interval shifts infinitely far to the left, leaving the limiting solution set $(2, \infty)$ .

Final answer

Solution set: $\left(-\infty, -\frac{1}{\epsilon}\right) \cup (2, \infty)$ . As $\epsilon \to 0^+$ , $-\frac{1}{\epsilon} \to -\infty$ , leaving the limiting solution set $(2, \infty)$ on the real line.

NEXT STEPS

Final Takeaways from Your Teacher

Practice isolating variables in linear equations involving symbolic parameters.
Verify your boundary points by substituting a specific small test value such as $\epsilon = 0.1$ to check if the denominator equals zero.
Review the asymptotic behavior of reciprocal functions $\frac{1}{\epsilon}$ as $\epsilon \to 0^+$ .

OverviewACEpath Teacher's Diagnosis

03 The student incorrectly assumes that the solution to a strictly positive rational inequality is the interior interval between the roots rather than the exterior union.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
Critical points: $x = 2$ and $x = -\frac{1}{\epsilon} \implies$ Since $\epsilon > 0$ , $-\frac{1}{\epsilon} < 2$
Step 2
The inequality is satisfied between the two roots $\implies$ Solution set : $(-\frac{1}{\epsilon}, 2)$
Correct
Sign testing: for $x \in (-\infty, -\frac{1}{\epsilon})$ , $\frac{(-)}{(-)} > 0$ ; for $x \in (-\frac{1}{\epsilon}, 2)$ , $\frac{(-)}{(+)} < 0$ ; for $x \in (2, \infty)$ , $\frac{(+)}{(+)} > 0$ . Therefore, the solution set is $(-\infty, -\frac{1}{\epsilon}) \cup (2, \infty)$ .
When $x$ is between $-\frac{1}{\epsilon}$ and $2$ , the numerator $x - 2$ is negative while the denominator $\epsilon x + 1$ is positive, making the quotient negative ( $< 0$ ). A rational expression with positive leading coefficients is positive outside the roots, not between them.
Step 3
As $\epsilon \to 0^+$ , the left boundary point $-\frac{1}{\epsilon}$ approaches $-\infty$ , while $2$ remains fixed.
Step 4
The shaded interval $(-\frac{1}{\epsilon}, 2)$ expands indefinitely to the left, covering $(-\infty, 2)$ in the limit.
Correct
The left interval $(-\infty, -\frac{1}{\epsilon})$ is pushed infinitely to the left and vanishes from any finite window, leaving $(2, \infty)$ as the limiting solution set.
Because you selected the interior interval in Step 2, your conclusion about the expanding shaded region followed your earlier mistake rather than the true exterior solution set.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly identified the critical values by setting the numerator and denominator to zero.

You properly ordered the critical values on the number line using the condition $\epsilon > 0$ .

You accurately found the limit of the boundary point $-\frac{1}{\epsilon}$ as $\epsilon \to 0^+$ .

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Sign Chart Region Misidentification

Assuming the solution to a strictly positive inequality lies between the critical values without testing signs in each interval.

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Conflating rational inequalities of the form $\frac{x - a}{x - b} > 0$ with inequalities where the product is negative ( $< 0$ ), overlooking that matching signs in the numerator and denominator yield positive values in the exterior intervals $(-\infty, a) \cup (b, \infty)$ .

REASONING PATH

Your ACEpath teacher’s suggested approach

Test Interval Signs: A quotient $\frac{A}{B} > 0$ requires $A$ and $B$ to have the same sign (both positive or both negative).
Pick a test point in each partitioned region or check the signs of the leading coefficients before choosing intervals.
Track Asymptotic Sets: If a boundary of an interval $(-\infty, c)$ approaches $-\infty$ , the interval gets pushed off the real line and vanishes in the limit.
Check whether the boundary is pushing the interval away or expanding it across the visible line.

CORRECTED PATH

ACEpath Teacher's Demonstration

Test the sign of $\frac{x - 2}{\epsilon x + 1}$ across the three regions partitioned by $-\frac{1}{\epsilon}$ and $2$ :
For $x < -\frac{1}{\epsilon}$ , both numerator and denominator are negative, so the quotient is positive ( $> 0$ ).
For $-\frac{1}{\epsilon} < x < 2$ , the numerator is negative and the denominator is positive, so the quotient is negative ( $< 0$ ).
For $x > 2$ , both numerator and denominator are positive, so the quotient is positive ( $> 0$ ).
Combine the positive regions to get the solution set $(-\infty, -\frac{1}{\epsilon}) \cup (2, \infty)$ .
Take the limit as $\epsilon \to 0^+$ : $-\frac{1}{\epsilon} \to -\infty$ , which pushes the left interval $(-\infty, -\frac{1}{\epsilon})$ entirely to $-\infty$ and leaves $(2, \infty)$ .

Final answer

Solution set: $(-\infty, -\frac{1}{\epsilon}) \cup (2, \infty)$ . In the limit $\epsilon \to 0^+$ , the left branch vanishes as $-\frac{1}{\epsilon} \to -\infty$ , leaving $(2, \infty)$ (an open circle at $x = 2$ shaded to $+\infty$ ).

NEXT STEPS

Final Takeaways from Your Teacher

Always construct a 3-interval sign chart or test a value like $x = 0$ to verify the sign between the roots.
Practice sketching the behavior of parameterized intervals as boundary parameters approach zero or infinity.

Lock it in

Question

Why is it mathematically invalid to multiply both sides of $\frac{x - 2}{x + 3} > 0$ by $x + 3$ ?

Answer

Because $x + 3$ can be negative (when $x < -3$ ). Multiplying by a negative quantity reverses the inequality direction, so you would lose part of the solution set.

Resources

Answer and steps

Final answer

Steps

Typical errors

01 The student clears the denominator by multiplying by εx + 1 without considering that it can be negative, losing the left interval (-∞, -1/ε).

Sample Student Work

Step-by-Step Review

WHAT YOU DID WELL

THE ERROR

WHY IT HAPPENED

REASONING PATH

CORRECTED PATH

NEXT STEPS

02 The student mistakenly solves εx + 1 = 0 as x = -ε instead of x = -1/ε, leading to an incorrect boundary point that vanishes rather than diverging as ε → 0⁺.

Sample Student Work

Step-by-Step Review

WHAT YOU DID WELL

THE ERROR

WHY IT HAPPENED

REASONING PATH

CORRECTED PATH

NEXT STEPS

03 The student incorrectly assumes that the solution to a strictly positive rational inequality is the interior interval between the roots rather than the exterior union.

Sample Student Work

Step-by-Step Review

WHAT YOU DID WELL

THE ERROR

WHY IT HAPPENED

REASONING PATH

CORRECTED PATH

NEXT STEPS

Lock it in

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