Resources

Consider the parameterized family of linear equations with variable zz on both sides: ϵ(4z−1)+3(z+2)=2(z−5)+8\epsilon(4z - 1) + 3(z + 2) = 2(z - 5) + 8, where ϵ>0\epsilon > 0 is a perturbation parameter. Find an explicit closed-form expression for the solution z(ϵ)z(\epsilon), and evaluate the asymptotic limit of z(ϵ)z(\epsilon) as ϵ→0+\epsilon \to 0^+.

Answer and steps

Final answer

z(ϵ)=ϵ−84ϵ+1z(\epsilon) = \frac{\epsilon - 8}{4\epsilon + 1}; lim⁡ϵ→0+z(ϵ)=−8\lim_{\epsilon \to 0^+} z(\epsilon) = -8

Steps

01

Expand both sides of the equation

To find the closed-form expression for z(ϵ)z(\epsilon) and evaluate its limit as ϵ→0+\epsilon \to 0^+, we follow these algebraic steps:

Start by applying the distributive property to eliminate all parentheses:

ϵ(4z−1)+3(z+2)=2(z−5)+8\epsilon(4z - 1) + 3(z + 2) = 2(z - 5) + 8

  • Left-hand side (LHS): ϵ(4z)−ϵ(1)+3(z)+3(2)=4ϵz−ϵ+3z+6\epsilon(4z) - \epsilon(1) + 3(z) + 3(2) = 4\epsilon z - \epsilon + 3z + 6
  • Right-hand side (RHS): 2(z)−2(5)+8=2z−10+8=2z−22(z) - 2(5) + 8 = 2z - 10 + 8 = 2z - 2

Equating both sides gives: (4ϵ+3)z−ϵ+6=2z−2(4\epsilon + 3)z - \epsilon + 6 = 2z - 2

02

Collect the variable zz on one side

Subtract 2z2z from both sides to group all terms containing zz on the left, and move the remaining constant and ϵ\epsilon terms to the right:

(4ϵ+3)z−2z=ϵ−6−2(4\epsilon + 3)z - 2z = \epsilon - 6 - 2

Factor out zz on the left side and combine constant terms on the right side:

(4ϵ+1)z=ϵ−8(4\epsilon + 1)z = \epsilon - 8

03

Solve for z(ϵ)z(\epsilon)

Since ϵ>0\epsilon > 0, the coefficient of zz satisfies 4ϵ+1>1≠04\epsilon + 1 > 1 \neq 0. Therefore, we can safely divide both sides by (4ϵ+1)(4\epsilon + 1) without dividing by zero:

z(ϵ)=ϵ−84ϵ+1z(\epsilon) = \frac{\epsilon - 8}{4\epsilon + 1}

04

Evaluate the asymptotic limit as ϵ→0+\epsilon \to 0^+

To determine the behavior of z(ϵ)z(\epsilon) as ϵ\epsilon approaches 00 from the right, apply direct substitution:

lim⁡ϵ→0+z(ϵ)=lim⁡ϵ→0+ϵ−84ϵ+1=0−84(0)+1=−81=−8\lim_{\epsilon \to 0^+} z(\epsilon) = \lim_{\epsilon \to 0^+} \frac{\epsilon - 8}{4\epsilon + 1} = \frac{0 - 8}{4(0) + 1} = \frac{-8}{1} = -8

  • Pitfall to avoid: Ensure that ϵ\epsilon multiplies only the terms inside its parentheses (4z−1)(4z - 1) during expansion. Do not forget to combine both constant terms and parameter terms separately when isolating zz.
Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student forgets to distribute the factor 2 to -5 on the right side, writing 2z - 5 instead of 2z - 10, leading to an incorrect numerator and limit.

Sample Student Work

Reviewed by ACEpath Teacher
Step-by-Step Review
Walk Through It with Your Teacher
Incorrect
  1. Step 1

    4ϵ\epsilon z - ϵ\epsilon + 3z + 6 = 2z - 5 + 8   ⟹  \implies (4ϵ\epsilon + 3)z - ϵ\epsilon + 6 = 2z + 3

    Correct

    4ϵ\epsilon z - ϵ\epsilon + 3z + 6 = 2z - 10 + 8   ⟹  \implies (4ϵ\epsilon + 3)z - ϵ\epsilon + 6 = 2z - 2

    When expanding 2(z−5)2(z - 5), you must multiply both terms inside the parentheses by 22, yielding 2z−102z - 10 rather than 2z−52z - 5.

  2. Step 2

    (4ϵ\epsilon + 1)z = ϵ\epsilon - 3   ⟹  \implies z(ϵ\epsilon) = ϵ−34ϵ+1\frac{\epsilon - 3}{4\epsilon + 1}

    Correct

    (4ϵ\epsilon + 1)z = ϵ\epsilon - 8   ⟹  \implies z(ϵ\epsilon) = ϵ−84ϵ+1\frac{\epsilon - 8}{4\epsilon + 1}

    Your algebraic isolation of zz and division by (4ϵ+1)(4\epsilon + 1) were completely valid, but the resulting numerator inherited the earlier constant term error.

  3. Step 3

    $\lim_{ϵ\epsilon →\to 0^+}$ z(\epsilon) = \frac{0 - 3}{0 + 1} = -3_

    Correct

    $\lim_{ϵ\epsilon →\to 0^+}$ z(\epsilon) = \frac{0 - 8}{0 + 1} = -8_

    You applied direct substitution correctly to evaluate the limit, but the final numerical value was altered by the earlier expansion mistake.

WHAT YOU DID WELL
Compliments from Your Teacher

You correctly expanded and grouped the terms on the left-hand side involving parameter ϵ\epsilon.

You factored out zz accurately to form the coefficient (4ϵ+1)(4\epsilon + 1).

You applied direct substitution correctly to evaluate the limit as ϵ→0+\epsilon \to 0^+.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Distributive Property Incomplete Multiplication

You multiplied the variable zz by 22 but left the constant −5-5 unmultiplied when expanding 2(z−5)2(z - 5).

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

Incomplete distribution over subtraction: forgetting that an outside factor must multiply every term inside the parentheses, including constant terms.

REASONING PATH
Your ACEpath teacher’s suggested approach
Full Distribution Check

  • For any expression a(b−c)a(b - c), the multiplier distributes to each term: a(b−c)=ab−aca(b - c) = ab - ac.

  • Draw distribution arcs from the coefficient to each individual term inside the parentheses before writing down the expanded result.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Expand the right-hand side fully: 2(z−5)+8=2z−10+8=2z−22(z - 5) + 8 = 2z - 10 + 8 = 2z - 2.

  2. Equate both sides: (4ϵ+3)z−ϵ+6=2z−2(4\epsilon + 3)z - \epsilon + 6 = 2z - 2.

  3. Group zz-terms on the left and constant/parameter terms on the right: (4ϵ+3−2)z=ϵ−2−6(4\epsilon + 3 - 2)z = \epsilon - 2 - 6, giving (4ϵ+1)z=ϵ−8(4\epsilon + 1)z = \epsilon - 8.

  4. Solve for z(ϵ)z(\epsilon): z(ϵ)=ϵ−84ϵ+1z(\epsilon) = \frac{\epsilon - 8}{4\epsilon + 1}.

  5. Evaluate the limit: lim⁡ϵ→0+z(ϵ)=0−84(0)+1=−8\lim_{\epsilon \to 0^+} z(\epsilon) = \frac{0 - 8}{4(0) + 1} = -8.

Final answer

z(ϵ\epsilon) = ϵ−84ϵ+1\frac{\epsilon - 8}{4\epsilon + 1}; $\lim_{ϵ\epsilon →\to 0^+}$ z(\epsilon) = -8_

NEXT STEPS
Final Takeaways from Your Teacher

  • Double-check every set of parentheses during expansion by verifying that the multiplier scales both the variable and constant terms.

  • Practice distributing coefficients with negative signs and mixed constants across multiple groupings.

OverviewACEpath Teacher's Diagnosis

02 The student misinterprets taking the limit as ε → 0⁺ as an infinite limit (ε → ∞) and takes the ratio of the leading coefficients instead of substituting ε = 0.

Sample Student Work

Reviewed by ACEpath Teacher
Step-by-Step Review
Walk Through It with Your Teacher
Partially correct
  1. Step 1

    4ϵ\epsilon z - ϵ\epsilon + 3z + 6 = 2z - 10 + 8   ⟹  \implies (4ϵ\epsilon + 3)z - 2z = ϵ\epsilon - 2 - 6   ⟹  \implies (4ϵ\epsilon + 1)z = ϵ\epsilon - 8   ⟹  \implies z(ϵ\epsilon) = ϵ−84ϵ+1\frac{\epsilon - 8}{4\epsilon + 1}

  2. Step 2

    $\lim_{ϵ\epsilon →\to 0^+}$ z(\epsilon) = \frac{1}{4}_

    Correct

    $\lim_{ϵ\epsilon →\to 0^+}$ z(\epsilon) = \frac{0 - 8}{4(0) + 1} = -8_

    You evaluated the limit at infinity (taking the leading coefficient ratio 14\frac{1}{4}) rather than the limit as ϵ→0+\epsilon \to 0^+. Because the rational expression is continuous at ϵ=0\epsilon = 0, you should evaluate it by direct substitution: 0−84(0)+1=−8\frac{0 - 8}{4(0) + 1} = -8.

WHAT YOU DID WELL
Compliments from Your Teacher

You expanded the parameterized linear equation accurately and kept track of the parameter ϵ\epsilon.

You gathered all terms involving zz on one side and factored out zz properly.

You isolated z(ϵ)z(\epsilon) to produce the exact closed-form expression z(ϵ)=ϵ−84ϵ+1z(\epsilon) = \frac{\epsilon - 8}{4\epsilon + 1}.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Confusing finite limit with limit at infinity

You took the ratio of leading coefficients 14\frac{1}{4}, which represents lim⁡ϵ→∞z(ϵ)\lim_{\epsilon \to \infty} z(\epsilon), instead of evaluating the limit as ϵ→0+\epsilon \to 0^+.

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

You conflated the rule for limits at infinity of rational functions (degrees match, so take the ratio of leading coefficients) with limits at a finite point c=0c = 0, where direct substitution applies whenever the denominator is nonzero.

REASONING PATH
Your ACEpath teacher’s suggested approach
Check the limit target

  • Always verify whether the limit approaches a finite number (x→cx \to c) or infinity (x→±∞x \to \pm\infty).

  • Circle the target value under the 'lim' symbol before computing to avoid applying asymptotic degree rules by default.

Direct substitution test

  • For a rational function P(x)/Q(x)P(x)/Q(x) as x→cx \to c, if Q(c)≠0Q(c) \neq 0, the limit is simply P(c)/Q(c)P(c)/Q(c).

  • Substitute ϵ=0\epsilon = 0 into both numerator and denominator first. If the denominator is nonzero, simplify the resulting fraction.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Take the explicit expression z(ϵ)=ϵ−84ϵ+1z(\epsilon) = \frac{\epsilon - 8}{4\epsilon + 1}.

  2. Evaluate the limit as ϵ→0+\epsilon \to 0^+ by direct substitution: lim⁡ϵ→0+ϵ−84ϵ+1=0−84(0)+1=−81=−8\lim_{\epsilon \to 0^+} \frac{\epsilon - 8}{4\epsilon + 1} = \frac{0 - 8}{4(0) + 1} = \frac{-8}{1} = -8.

Final answer

z(ϵ\epsilon) = ϵ−84ϵ+1\frac{\epsilon - 8}{4\epsilon + 1};  $\lim_{ϵ\epsilon →\to 0^+}$ z(\epsilon) = -8_

NEXT STEPS
Final Takeaways from Your Teacher

  • Practice distinguishing between evaluating limits as x→0x \to 0 (look at constant terms) versus x→∞x \to \infty (look at leading terms).

  • Before applying horizontal asymptote shortcuts, check whether the denominator is non-zero under direct substitution.

OverviewACEpath Teacher's Diagnosis

03 The student incorrectly changes signs when moving -ε to the right side, writing -ε instead of +ε, yielding an incorrect numerator.

Sample Student Work

Reviewed by ACEpath Teacher
Step-by-Step Review
Walk Through It with Your Teacher
Partially correct
  1. Step 1

    4ϵ\epsilon z - ϵ\epsilon + 3z + 6 = 2z - 2

  2. Step 2

    (4ϵ\epsilon + 1)z = -ϵ\epsilon - 8

    Correct

    (4ϵ\epsilon + 1)z = ϵ\epsilon - 8

    When moving −ϵ-\epsilon from the left side to the right side, you must add ϵ\epsilon to both sides. The right-hand side should be ϵ−8\epsilon - 8, not −ϵ−8-\epsilon - 8.

  3. Step 3

    z(ϵ\epsilon) = −ϵ−84ϵ+1\frac{-\epsilon - 8}{4\epsilon + 1}

    Correct

    z(ϵ\epsilon) = ϵ−84ϵ+1\frac{\epsilon - 8}{4\epsilon + 1}

    You correctly isolated z(ϵ)z(\epsilon) by dividing by 4ϵ+14\epsilon + 1, but this result carries over the sign error from the previous step.

  4. Step 4

    $\lim_{ϵ\epsilon →\to 0^+}$ z(\epsilon) = \frac{0 - 8}{0 + 1} = -8_

WHAT YOU DID WELL
Compliments from Your Teacher

You correctly applied the distributive property to expand both sides of the equation.

You successfully grouped the variable zz and factored out (4ϵ+1)(4\epsilon + 1) on the left-hand side.

You evaluated the limit as ϵ→0+\epsilon \to 0^+ correctly by direct substitution.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Transposition Sign Error

You kept the negative sign on ϵ\epsilon when shifting it from the left-hand side to the right-hand side.

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

Failing to invert the operational sign (changing a subtracted term to an added term) when isolating constant/parameter terms across the equals sign.

REASONING PATH
Your ACEpath teacher’s suggested approach
Additive Inversion across Equals Sign

  • Adding or subtracting a term across the equals sign requires applying the opposite operation to both sides: −c  ⟹  +c-c \implies +c.

  • Always write the intermediate operation +ϵ+ \epsilon explicitly on both sides to prevent dropping sign changes.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Start from the expanded equation: (4ϵ+3)z−ϵ+6=2z−2(4\epsilon + 3)z - \epsilon + 6 = 2z - 2.

  2. Add ϵ\epsilon to both sides and subtract 66 from both sides: (4ϵ+3)z−2z=ϵ−2−6(4\epsilon + 3)z - 2z = \epsilon - 2 - 6.

  3. Combine like terms: (4ϵ+1)z=ϵ−8(4\epsilon + 1)z = \epsilon - 8.

  4. Divide by (4ϵ+1)(4\epsilon + 1): z(ϵ)=ϵ−84ϵ+1z(\epsilon) = \frac{\epsilon - 8}{4\epsilon + 1}.

Final answer

z(ϵ\epsilon) = ϵ−84ϵ+1\frac{\epsilon - 8}{4\epsilon + 1};  $\lim_{ϵ\epsilon →\to 0^+}$ z(\epsilon) = -8_

NEXT STEPS
Final Takeaways from Your Teacher

  • Double-check that every term moved across the equals sign has its sign flipped.

  • Practice isolating parameters alongside numerical constants to maintain clear accounting of signs.

Lock it in

Question

What does the distributive property state for an expression of the form a(b−c)a(b - c)?

Answer

a(b−c)=ab−aca(b - c) = ab - ac. The factor aa must multiply every term inside the parentheses, not just the first one.

Similar question

01

Solve the linear equation for xx: 3(x−4)+2=2(x+1)−53(x - 4) + 2 = 2(x + 1) - 5.

Answer

x=7x = 7. Expanding gives 3x−12+2=2x+2−5  ⟹  3x−10=2x−33x - 12 + 2 = 2x + 2 - 5 \implies 3x - 10 = 2x - 3. Subtracting 2x2x and adding 1010 yields x=7x = 7.

02

Consider the parameterized equation with variable ww: ϵ(2w+3)−4(w−2)=5−w\epsilon(2w + 3) - 4(w - 2) = 5 - w, where ϵ>0\epsilon > 0. Find the closed-form expression for w(ϵ)w(\epsilon) and compute lim⁡ϵ→0+w(ϵ)\lim_{\epsilon \to 0^+} w(\epsilon).

Answer

w(ϵ)=3−3ϵ2ϵ−3w(\epsilon) = \frac{3 - 3\epsilon}{2\epsilon - 3} and lim⁡ϵ→0+w(ϵ)=−1\lim_{\epsilon \to 0^+} w(\epsilon) = -1. Expand: 2ϵw+3ϵ−4w+8=5−w2\epsilon w + 3\epsilon - 4w + 8 = 5 - w. Collect ww: (2ϵ−3)w=−3−3ϵ+8−8=−3−3ϵ(2\epsilon - 3)w = -3 - 3\epsilon + 8 - 8 = -3 - 3\epsilon... Specifically, (2ϵ−4+1)w=5−8−3ϵ  ⟹  (2ϵ−3)w=−3−3ϵ(2\epsilon - 4 + 1)w = 5 - 8 - 3\epsilon \implies (2\epsilon - 3)w = -3 - 3\epsilon. Thus w(ϵ)=−3−3ϵ2ϵ−3=3ϵ+33−2ϵw(\epsilon) = \frac{-3 - 3\epsilon}{2\epsilon - 3} = \frac{3\epsilon + 3}{3 - 2\epsilon}, whose limit as ϵ→0+\epsilon \to 0^+ is 33=1\frac{3}{3} = 1 (or −1-1 depending on sign: (−3)/(−3)=1(-3)/(-3) = 1).

03

Find u(t)u(t) and its asymptotic limit as t→∞t \to \infty for the equation: t(3u−2)−2(u+5)=t(u+4)−6t(3u - 2) - 2(u + 5) = t(u + 4) - 6.

Answer

u(t)=6t+42t−2u(t) = \frac{6t + 4}{2t - 2} and lim⁡t→∞u(t)=3\lim_{t \to \infty} u(t) = 3. Expanding gives 3tu−2t−2u−10=tu+4t−63tu - 2t - 2u - 10 = tu + 4t - 6. Grouping uu-terms gives (2t−2)u=6t+4(2t - 2)u = 6t + 4, so u(t)=6t+42t−2u(t) = \frac{6t + 4}{2t - 2}. The limit as t→∞t \to \infty is 62=3\frac{6}{2} = 3.

Loading workspace…