Home›Calculus 1›Limits at infinity›Problem 1.22

Limit of \( \displaystyle \frac{4 x^{2} - 1}{2 x^{2} + x} \) as \( x \to \infty \)

Problem 1.22 · medium Video

Evaluate \( \displaystyle \lim_{x \to \infty} \frac{4 x^{2} - 1}{2 x^{2} + x} \).

\[ \lim_{x \to \infty}\left(\frac{4 x^{2} - 1}{2 x^{2} + x}\right) \]
limit algebraStart with the original limit expression. Divide both numerator and denominator by x**2.✓ Proved
\[ = \lim_{x \to \infty}\left(\frac{4 - \frac{1}{x^{2}}}{2 + \frac{1}{x}}\right) \]
algebraSimplify the fractions inside the limit.✓ Proved
\[ = \lim_{x \to \infty}\left(4 - \frac{1}{x^{2}}\right) \left(\lim_{x \to \infty}\left(2 + \frac{1}{x}\right)\right)^{-1} \]
limit-lawApply the quotient rule for limits.✓ Proved
\[ = 2 \]
limit simplifyEvaluate the limits of the numerator and denominator separately. Final simplification.✓ Proved

Answer \( 2 \)

The graph of f(x) = (4x² - 1)/(2x² + x). A point runs right along the curve; its height settles toward 2.

✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. How we verify

The full receipt

Line	Status	Checked by	Detail
1	✓ Proved	sympy 1.14.0	line 1 is the problem as stated
2	✓ Proved	sympy 1.14.0	simplify(a - b) reduced to 0 undefined where 2x*2 + x = 0
3	✓ Proved	sympy 1.14.0	simplify(a - b) reduced to 0 undefined where 2x*2 + x = 0 undefined where 2 + 1/x = 0 undefined where x = 0
4	✓ Proved	sympy 1.14.0	simplify(a - b) reduced to 0 undefined where 2 + 1/x = 0 undefined where x = 0 undefined where Limit(2 + 1/x, x, oo, dir='-') = 0
5	✓ Proved	sympy 1.14.0	simplify(a - b) reduced to 0 undefined where x = 0 undefined where Limit(2 + 1/x, x, oo, dir='-') = 0
6	✓ 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: pass
qwen3.6:27b-mlx: pass — The solution correctly applies algebraic manipulation and limit laws to evaluate the limit at infinity. Each step changes only one aspect of the expression and uses valid labels from the fixed vocabulary.

Senior review claude-sonnet-5-5, 2026-10-03: pass — The solution is correct. Dividing by x**2 is algebra, the quotient limit law is applied properly, the limits of 4-1/x**2 and 2+1/x are 4 and 2, and 4/2 = 2 matches the stated answer. SymPy proved every line. The only panel objection is unspecified, so there is nothing to rule on, and no lines are weak.

Every verdict on record (7)

qwen3.6:27b-mlx: pass 2026-10-03 — The solution correctly applies algebraic manipulation and limit laws to evaluate the limit at infinity. Each step changes only one aspect of the expression and uses valid labels from the fixed vocabulary.
gpt-oss:20b: pass 2026-10-03
claude-sonnet-5-5: pass 2026-10-03 — The solution is correct. Dividing by x**2 is algebra, the quotient limit law is applied properly, the limits of 4-1/x**2 and 2+1/x are 4 and 2, and 4/2 = 2 matches the stated answer. SymPy proved every line. The only panel objection is unspecified, so there is nothing to rule on, and no lines are weak.
qwen3.6:27b-mlx: pass 2026-09-29
gpt-oss:20b: pass 2026-09-29
qwen3.6:27b-mlx: fail (error) 2026-09-29 — Step 5 claims to evaluate the limits of the numerator and denominator separately, but it skips the intermediate step of evaluating the individual limits (e.g., Limit(4 - 1/x**2, x, oo) -> 4). It jumps directly from the limit expressions to the constants, violating the 'one thing per step' rule by combining limit evaluation and substitution.
gpt-oss:20b: pass 2026-09-29

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 2026-10-03 with SymPy 1.14.0.