Limit of \( \displaystyle \frac{4 x^{2} + 2 x - 6}{2 x - 2} \) as \( x \to 1 \)
Problem 1.312 · medium
Evaluate \( \displaystyle \lim_{x \to 1} \frac{4 x^{2} + 2 x - 6}{2 x - 2} \).
- \[ \lim_{x \to 1^+}\left(\frac{4 x^{2} + 2 x - 6}{2 x - 2}\right) \]limitIdentify the limit to evaluate.✓ Proved
- \[ = \lim_{x \to 1^+}\left(\frac{\left(2 x - 3\right) \left(2 x + 2\right)}{2 x - 2}\right) \]factorFactor the numerator.Not checked
- \[ = \lim_{x \to 1^+}\left(\frac{\frac{d}{d x} \left(4 x^{2} + 2 x - 6\right)}{\frac{d}{d x} \left(2 x - 2\right)}\right) \]lhopitalApply L'Hopital's rule since the limit is an indeterminate form 0/0.Not checked
- \[ = \lim_{x \to 1^+}\left(4 x + 1\right) \]simplifyDifferentiate the numerator and the denominator.✓ Proved
- \[ = 5 \]limitEvaluate the limit by substituting x = 1.✓ Proved
Answer \( 5 \)
Lines: 4 proved, 2 not checked. The answer was also checked a second way, without looking at the solution. A reviewer's objection to the explanation stands; see the receipt.
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | ✓ Proved | sympy 1.14.0 | line 1 is the problem as stated |
| 2 | Not checked | claude-sonnet-5-5 script, run by sympy 1.14.0 | claude-sonnet-5-5 script, run by sympy 1.14.0: expanded numerator difference (line1 - line2) = 4*x, denominator difference = 0; factorization is wrong (correct is (2x+3)(2x-2)) (checks/1.312/line2-6f5ec7d0.py)undefined where 2*x - 2 = 0 |
| 3 | Not checked | claude-sonnet-5-5 script, run by sympy 1.14.0 | claude-sonnet-5-5 script, run by sympy 1.14.0: L-Hopital applied to line 2 numerator/denominator gives 4*x - 1 but line 3 has 4*x + 1; difference -2 (checks/1.312/line3-bfd37aef.py)undefined where 2*x - 2 = 0 undefined where Derivative(2*x - 2, x) = 0 |
| 4 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 undefined where Derivative(2*x - 2, x) = 0 |
| 5 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 |
| answer | ✓ Proved | sympy 1.14.0 | final line against the stated answer: simplify(a - b) reduced to 0 |
| answer, a second way | ✓ Proved | sympy 1.14.0 | SymPy took the limit from both sides and got the stated value |
Reviewers
gpt-oss:20b: fail (error) — Step 2 incorrectly factors the numerator: 4*x**2 + 2*x - 6 does not equal (2*x - 3)*(2*x + 2). This violates the rule that each step must change only one thing and apply a correct algebraic transformation.qwen3.6:27b-mlx: fail (error) — The solution mixes factoring and L'Hopital's rule inconsistently. Step 2 factors the numerator but leaves the denominator unfactored, and Step 3 applies L'Hopital's rule to the expression from Step 2, which is algebraically equivalent to the original but structurally different; however, the primary defect is that Step 3 claims to apply L'Hopital's rule to the expression in Step 2, but L'Hopital's rule applies to the original limit form, not necessarily to a partially factored form in this context without re-verifying the 0/0 condition on the specific expression shown (though it holds). More critically, Step 3 introduces 'Derivative' objects, and Step 4 simplifies them. The transition from Step 2 to Step 3 is logically disjointed: one does not typically factor then immediately switch to L'Hopital's on the factored form in a single step chain without justification, but the real error is that Step 3's expression `Derivative(4*x**2 + 2*x - 6, x)/Derivative(2*x - 2, x)` is not algebraically equal to Step 2's expression `(2*x - 3)*(2*x + 2)/(2*x - 2)`. The prompt states a CAS confirmed consecutive lines are algebraically equal, but `Derivative(...)` is not algebraically equal to the polynomial expression unless the derivative is evaluated. If the CAS checked equality of the *values* of the limits, that's fine, but the steps themselves are not algebraically equal expressions. Wait, the prompt says 'A computer algebra system has already confirmed that consecutive lines are algebraically equal'. This implies I should not check algebra. Let's look at the rules. Step 2 factors the numerator. Step 3 applies L'Hopital. L'Hopital's rule is a limit law, not an algebraic manipulation. The expression in Step 3 is the result of applying L'Hopital's rule to the limit in Step 1 (or 2). But Step 2 changed the expression. If Step 3 is derived from Step 2, it's applying L'Hopital to `(2*x - 3)*(2*x + 2)/(2*x - 2)`. The derivative of the numerator `(2*x - 3)*(2*x + 2)` is `2*(2*x+2) + (2*x-3)*2 = 4x + 4 + 4x - 6 = 8x - 2`. The derivative of the denominator is `2`. So the ratio is `(8x-2)/2`. Step 4 shows `(8*x + 2)/2`. This is `(8x+2)/2 = 4x+1`. At x=1, this is 5. The correct derivative of the numerator `4x^2+2x-6` is `8x+2`. The derivative of the factored numerator `(2x-3)(2x+2)` is `8x-2`. Step 3 writes `Derivative(4*x**2 + 2*x - 6, x)`, which is the derivative of the *original* numerator, not the factored one from Step 2. Thus, Step 3 is not a valid application of L'Hopital's rule to the expression in Step 2. It jumps back to the original form's derivative. This is a logic error in the step chain.
Senior review claude-sonnet-5-5, 2026-10-06: fail — Line 2 is false: (2x-3)(2x+2) expands to 4x^2-2x-6, not 4x^2+2x-6, and it equals -4 at x=1 (so the limit would be infinite). The correct factorization is (2x+3)(2x-2). Line 3 is a valid L'Hopital step from line 1 but does not follow from line 2, so the final answer 5 is right while the written chain is invalid.
qwen3.6:27b-mlx: uphold — Line 3 differentiates the unfactored numerator 4x^2+2x-6, so it follows from line 1 and not from line 2, and the chain is broken; the panel's stated reason is muddled, since the real defect is that line 2 is itself false.
Every verdict on record (5)
gpt-oss:20b: fail (error) 2026-10-06 — Step 2 incorrectly factors the numerator: 4*x**2 + 2*x - 6 does not equal (2*x - 3)*(2*x + 2). This violates the rule that each step must change only one thing and apply a correct algebraic transformation.qwen3.6:27b-mlx: fail (error) 2026-10-06 — The solution mixes factoring and L'Hopital's rule inconsistently. Step 2 factors the numerator but leaves the denominator unfactored, and Step 3 applies L'Hopital's rule to the expression from Step 2, which is algebraically equivalent to the original but structurally different; however, the primary defect is that Step 3 claims to apply L'Hopital's rule to the expression in Step 2, but L'Hopital's rule applies to the original limit form, not necessarily to a partially factored form in this context without re-verifying the 0/0 condition on the specific expression shown (though it holds). More critically, Step 3 introduces 'Derivative' objects, and Step 4 simplifies them. The transition from Step 2 to Step 3 is logically disjointed: one does not typically factor then immediately switch to L'Hopital's on the factored form in a single step chain without justification, but the real error is that Step 3's expression `Derivative(4*x**2 + 2*x - 6, x)/Derivative(2*x - 2, x)` is not algebraically equal to Step 2's expression `(2*x - 3)*(2*x + 2)/(2*x - 2)`. The prompt states a CAS confirmed consecutive lines are algebraically equal, but `Derivative(...)` is not algebraically equal to the polynomial expression unless the derivative is evaluated. If the CAS checked equality of the *values* of the limits, that's fine, but the steps themselves are not algebraically equal expressions. Wait, the prompt says 'A computer algebra system has already confirmed that consecutive lines are algebraically equal'. This implies I should not check algebra. Let's look at the rules. Step 2 factors the numerator. Step 3 applies L'Hopital. L'Hopital's rule is a limit law, not an algebraic manipulation. The expression in Step 3 is the result of applying L'Hopital's rule to the limit in Step 1 (or 2). But Step 2 changed the expression. If Step 3 is derived from Step 2, it's applying L'Hopital to `(2*x - 3)*(2*x + 2)/(2*x - 2)`. The derivative of the numerator `(2*x - 3)*(2*x + 2)` is `2*(2*x+2) + (2*x-3)*2 = 4x + 4 + 4x - 6 = 8x - 2`. The derivative of the denominator is `2`. So the ratio is `(8x-2)/2`. Step 4 shows `(8*x + 2)/2`. This is `(8x+2)/2 = 4x+1`. At x=1, this is 5. The correct derivative of the numerator `4x^2+2x-6` is `8x+2`. The derivative of the factored numerator `(2x-3)(2x+2)` is `8x-2`. Step 3 writes `Derivative(4*x**2 + 2*x - 6, x)`, which is the derivative of the *original* numerator, not the factored one from Step 2. Thus, Step 3 is not a valid application of L'Hopital's rule to the expression in Step 2. It jumps back to the original form's derivative. This is a logic error in the step chain.claude-sonnet-5-5: fail (error) 2026-10-06 — Line 2 is false: (2x-3)(2x+2) expands to 4x^2-2x-6, not 4x^2+2x-6, and it equals -4 at x=1 (so the limit would be infinite). The correct factorization is (2x+3)(2x-2). Line 3 is a valid L'Hopital step from line 1 but does not follow from line 2, so the final answer 5 is right while the written chain is invalid.qwen3.6:27b-mlx: fail (error) 2026-10-06 — The solution mixes factoring and L'Hopital's rule. Step 2 factors the numerator, but Step 3 applies L'Hopital's rule to the original expression (or implies it on the factored form without simplifying), which is logically inconsistent with the previous step. Specifically, if one factors, one should cancel; if one uses L'Hopital, one differentiates. The transition from Step 2 to Step 3 is a non-sequitur because Step 2 changed the expression to a factored form, but Step 3 reverts to a derivative form of the original numerator/denominator without showing the differentiation step on the factored expression or canceling first. Furthermore, Step 3 claims to apply L'Hopital's rule, but the expression in Step 3 is `Limit(Derivative(...)/Derivative(...), x, 1)`. This is not the result of applying L'Hopital's rule to the expression in Step 2 (`(2*x - 3)*(2*x + 2)/(2*x - 2)`). The derivative of the numerator in Step 2 would be `2*(2*x+2) + (2*x-3)*2 = 4*x+4+4*x-6 = 8*x-2`, not `8*x+2`. Wait, the derivative of `4*x**2 + 2*x - 6` is `8*x + 2`. So Step 3 is applying L'Hopital to the *original* expression, ignoring the factorization in Step 2. This violates the 'one step changes one thing' rule because Step 2 changed the expression, and Step 3 ignored that change to apply a rule to the previous state, or it implies a jump. Actually, the bigger issue is that Step 2 is unnecessary and misleading if Step 3 is L'Hopital on the original. But strictly, Step 3's input is the output of Step 2. The output of Step 2 is `(2*x - 3)*(2*x + 2)/(2*x - 2)`. Applying L'Hopital to THIS expression requires differentiating the product. The step shows `Derivative(4*x**2 + 2*x - 6, x)`, which is the derivative of the *unfactored* numerator. Thus, Step 3 does not follow from Step 2. It follows from Step 1. This is a logical error in the chain of reasoning.gpt-oss:20b: pass 2026-10-06
Proved: SymPy reduced the difference between the two sides to zero. Checked independently: a separate
method, named above, confirmed it. Checked numerically: the two sides agree at every sampled point, which is evidence,
not proof. Reviewed: a model or a person read it; that is all a sentence can have. Solution by
gemma4:26b, checked 2026-10-06 with SymPy 1.14.0.