Home›Calculus 3›Limits of functions of two variables›Problem 10.596

Limits of functions of two variables

Problem 10.596 · medium

Find \( \displaystyle \lim_{(x,y)\to(-1,1)} \frac{x^{2} + 2 x y + y^{2}}{x + y} \), or show it does not exist.

Substitution gives 0/0. Factor the numerator and cancel the common factor (it is nonzero near, but not at, the point).
\[ \left(x + y\right)^{2} = x^{2} + 2 x y + y^{2} \]
The numerator factored.✓ Proved
\[ \left. x + y \right|_{\substack{ x=-1\\ y=1 }} = 0 \]
Then substitute.✓ Proved

Answer \( 0 \)

Lines: 2 proved, 1 not checked. The answer was also checked a second way, without looking at the solution. The reviewers disagree about how one step is explained; every verdict is in the receipt. How we verify

The full receipt

Line	Status	Checked by	Detail
1	Not checked	—	a sentence; read, not computed
2	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
3	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
answer, a second way	✓ Checked independently	sympy 1.14.0 + mpmath 1.3.0	values along ten paths into the point (lines and parabolas) all approach it

Reviewers

gpt-oss:20b: pass
qwen3.6:27b-mlx: fail (style) — [domain objection, downgraded to style] The solution claims the limit is 0, but the algebraic simplification yields x+y, which approaches 0 as (x,y) approaches (-1,1). The stated answer is correct, but the solution text is incomplete and misleading because it stops at the substitution step without explicitly stating the final limit value or concluding the argument. More critically, the prompt asks to review the *solution lines provided*. Line 1 says 'Factor... and cancel', Line 2 shows the factorization, Line 3 shows substitution into the factor (x+y) yielding 0. The logic is sound: (x+y)^2/(x+y) = x+y -> 0. Wait, let me re-read carefully. Line 3 says 'Subs(x + y, (x, y), (-1, 1)) = 0'. This is the value of the simplified expression. The limit is indeed 0. Is there an error? The problem asks for the limit. The solution derives that the simplified function is x+y and its value at the point is 0. Since the simplified function is continuous, the limit is 0. The steps are logically sound. Why did I think it was fail? Let me check the 'Stated answer: 0'. The limit is 0. The solution shows the numerator factors to (x+y)^2. Canceling (x+y) gives x+y. The limit of x+y as (x,y)->(-1,1) is 0. The solution lines provided are: 1. Sentence describing the method. 2. Equation showing factorization. 3. Equation showing the result of substituting into the simplified term. This seems correct. However, line 3 is labeled 'proved' by a CAS. The CAS checked 'Subs(x + y, (x, y), (-1, 1)) = 0'. This is trivially true. The sentence in line 1 says 'cancel the common factor...'. It does not explicitly write the simplified expression, but line 3 implies the simplified expression is x+y. Is it misleading? A student might not see where x+y came from if they don't connect line 2 and 3. But line 2 is just the expansion of the numerator. It doesn't show the division. Line 3 substitutes into 'x+y'. Where did 'x+y' come from? It's the denominator, or the factor. The cancellation step is described in text but not shown in an equation. This is a 'style' or 'misleading' issue because the algebraic step of cancellation is skipped in the equations. But is it an error? The prompt asks to find what is wrong. The solution is technically correct in its conclusion. But is it 'sound'? Line 1 says 'cancel...'. Line 3 evaluates 'x+y'. It assumes the reader knows the simplified form is x+y. This is a bit of a leap. Let's look closer. Line 2: (x+y)^2 = ... Line 3: Subs(x+y, ...) = 0. The connection is that the fraction simplifies to x+y. This is not explicitly stated as an equation. Is this an error? No. Is it misleading? Maybe. But usually, these reviews look for mathematical errors. Let's look at the limit again. Limit of (x+y)^2/(x+y) is limit of x+y. As (x,y)->(-1,1), x+y -> 0. The answer is 0. The solution is correct. Why would it be fail? Maybe the 'Stated answer' is wrong? No, 0 is correct. Maybe the solution is incomplete? It doesn't explicitly state 'Therefore the limit is 0'. It just ends with the substitution. This is a 'style' issue or 'misleading' because it doesn't conclude. However, often 'pass' is reserved for fully rigorous solutions. Let's look for a subtle error. Is the factor nonzero near the point? x+y=0 is the line y=-x. The point (-1,1) is on this line. So the function is undefined on the entire line passing through the point. The limit exists if all paths give the same value. Since it simplifies to x+y everywhere except on the line x+y=0, and x+y is continuous, the limit is indeed 0. The text says 'it is nonzero near, but not at, the point'. This is true for any neighborhood excluding the line itself? No, 'near' usually implies a punctured disk. In any punctured disk around (-1,1), there are points where x+y=0 (the line passes through the point). So the factor IS zero at points arbitrarily close to (-1,1). The statement 'it is nonzero near... the point' is FALSE. The factor x+y is zero on the line y=-x, which passes through (-1,1). Therefore, in any neighborhood of (-1,1), there are points where the denominator is zero. The cancellation is valid for points where x+y != 0. The limit definition requires approaching the point through points in the domain. The domain excludes the line x+y=0. The simplified function x+y agrees with the original function on the domain. Since x+y is continuous, the limit is the value of x+y at the point. The text 'it is nonzero near... the point' is technically incorrect because it IS zero at points arbitrarily close to the point (just not AT the point? No, it's zero on a line passing through the point, so it's zero at points arbitrarily close). The phrase 'nonzero near... the point' suggests there is a neighborhood where it is nonzero, which is false. It is zero on a set that accumulates at the point. This is a significant conceptual error in the justification.

Every verdict on record (6)

qwen3.6:27b-mlx: fail (style) 2026-10-17 — [domain objection, downgraded to style] The solution claims the limit is 0, but the algebraic simplification yields x+y, which approaches 0 as (x,y) approaches (-1,1). The stated answer is correct, but the solution text is incomplete and misleading because it stops at the substitution step without explicitly stating the final limit value or concluding the argument. More critically, the prompt asks to review the *solution lines provided*. Line 1 says 'Factor... and cancel', Line 2 shows the factorization, Line 3 shows substitution into the factor (x+y) yielding 0. The logic is sound: (x+y)^2/(x+y) = x+y -> 0. Wait, let me re-read carefully. Line 3 says 'Subs(x + y, (x, y), (-1, 1)) = 0'. This is the value of the simplified expression. The limit is indeed 0. Is there an error? The problem asks for the limit. The solution derives that the simplified function is x+y and its value at the point is 0. Since the simplified function is continuous, the limit is 0. The steps are logically sound. Why did I think it was fail? Let me check the 'Stated answer: 0'. The limit is 0. The solution shows the numerator factors to (x+y)^2. Canceling (x+y) gives x+y. The limit of x+y as (x,y)->(-1,1) is 0. The solution lines provided are: 1. Sentence describing the method. 2. Equation showing factorization. 3. Equation showing the result of substituting into the simplified term. This seems correct. However, line 3 is labeled 'proved' by a CAS. The CAS checked 'Subs(x + y, (x, y), (-1, 1)) = 0'. This is trivially true. The sentence in line 1 says 'cancel the common factor...'. It does not explicitly write the simplified expression, but line 3 implies the simplified expression is x+y. Is it misleading? A student might not see where x+y came from if they don't connect line 2 and 3. But line 2 is just the expansion of the numerator. It doesn't show the division. Line 3 substitutes into 'x+y'. Where did 'x+y' come from? It's the denominator, or the factor. The cancellation step is described in text but not shown in an equation. This is a 'style' or 'misleading' issue because the algebraic step of cancellation is skipped in the equations. But is it an error? The prompt asks to find what is wrong. The solution is technically correct in its conclusion. But is it 'sound'? Line 1 says 'cancel...'. Line 3 evaluates 'x+y'. It assumes the reader knows the simplified form is x+y. This is a bit of a leap. Let's look closer. Line 2: (x+y)^2 = ... Line 3: Subs(x+y, ...) = 0. The connection is that the fraction simplifies to x+y. This is not explicitly stated as an equation. Is this an error? No. Is it misleading? Maybe. But usually, these reviews look for mathematical errors. Let's look at the limit again. Limit of (x+y)^2/(x+y) is limit of x+y. As (x,y)->(-1,1), x+y -> 0. The answer is 0. The solution is correct. Why would it be fail? Maybe the 'Stated answer' is wrong? No, 0 is correct. Maybe the solution is incomplete? It doesn't explicitly state 'Therefore the limit is 0'. It just ends with the substitution. This is a 'style' issue or 'misleading' because it doesn't conclude. However, often 'pass' is reserved for fully rigorous solutions. Let's look for a subtle error. Is the factor nonzero near the point? x+y=0 is the line y=-x. The point (-1,1) is on this line. So the function is undefined on the entire line passing through the point. The limit exists if all paths give the same value. Since it simplifies to x+y everywhere except on the line x+y=0, and x+y is continuous, the limit is indeed 0. The text says 'it is nonzero near, but not at, the point'. This is true for any neighborhood excluding the line itself? No, 'near' usually implies a punctured disk. In any punctured disk around (-1,1), there are points where x+y=0 (the line passes through the point). So the factor IS zero at points arbitrarily close to (-1,1). The statement 'it is nonzero near... the point' is FALSE. The factor x+y is zero on the line y=-x, which passes through (-1,1). Therefore, in any neighborhood of (-1,1), there are points where the denominator is zero. The cancellation is valid for points where x+y != 0. The limit definition requires approaching the point through points in the domain. The domain excludes the line x+y=0. The simplified function x+y agrees with the original function on the domain. Since x+y is continuous, the limit is the value of x+y at the point. The text 'it is nonzero near... the point' is technically incorrect because it IS zero at points arbitrarily close to the point (just not AT the point? No, it's zero on a line passing through the point, so it's zero at points arbitrarily close). The phrase 'nonzero near... the point' suggests there is a neighborhood where it is nonzero, which is false. It is zero on a set that accumulates at the point. This is a significant conceptual error in the justification.
gpt-oss:20b: pass 2026-10-17
qwen3.6:27b-mlx: fail (error) 2026-10-15 — The solution fails to compute the limit. It correctly factors the numerator and identifies the indeterminate form, but stops after substituting into the factor (x+y) to show it is zero, rather than simplifying the expression to (x+y) and evaluating the limit as 0.
gpt-oss:20b: pass 2026-10-15
qwen3.6:27b-mlx: fail (error) 2026-10-14 — The solution fails to compute the limit. It correctly factors the numerator and identifies the indeterminate form, but it stops after substituting into the factor (x+y) to get 0, rather than simplifying the expression to (x+y) and then evaluating the limit, which is 0. As written, it does not show how the limit is derived from the factored form.
gpt-oss:20b: pass 2026-10-14

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 generator:structured/multivariable_limits, checked 2026-10-17 with SymPy 1.14.0.