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πŸ“ General vs path limits multivariable (13 MCQs)

πŸ“– From Calculus β€’ 14. Partial Derivatives Calculus β€’ 13 questions available

What is General vs path limits multivariable?

Definition:
Path limits check specific trajectories, while the general limit requires consistency across ALL possible approaches simultaneously within a neighborhood.

Example:
f(x,y)=x2yx4+y2f(x,y) = \frac{x^2y}{x^4+y^2} has path limit 0 along lines y=mxy=mx but fails generally because along y=x2y=x^2 the limit is 1/21/2.

Reason:
Distinguishing these prevents false conclusions; matching path limits suggest but do not prove existence, necessitating squeeze theorem or polar conversion for confirmation.

3
Easy
5
Medium
5
Hard

πŸ“ All General vs path limits multivariable MCQs

Q1. A student claims that since lim⁑(x,y)β†’(0,0)f(x,y)=L\lim_{(x,y)\to(0,0)} f(x,y) = L along every straight line through the origin, the general limit must exist and equal LL. Which statement best evaluates this reasoning?

A.The reasoning is valid because lines form a basis for all smooth curves near the origin.
B.The reasoning is flawed; agreement along lines does not guarantee agreement along parabolic or higher-order paths. βœ…
C.The reasoning is valid only if ff is differentiable at the origin.
D.The reasoning is flawed unless the function is defined piecewise.
πŸ’‘ Difficulty: medium | βœ… Correct: B

πŸ“– Explanation: Agreement along all straight lines is necessary but insufficient for the existence of a general multivariable limit. A counterexample such as f(x,y)=x2yx4+y2f(x,y)=\frac{x^2y}{x^4+y^2} yields zero along every line yet approaches different values along y=x2y=x^2, demonstrating that smooth nonlinear curves can reveal path dependence invisible to linear analysis.

Q2. Consider f(x,y)=x3x2+y2f(x,y) = \frac{x^3}{x^2 + y^2} for (x,y)β‰ (0,0)(x,y) \neq (0,0). If you test limits along all smooth curves parameterized by x=t,y=g(t)x=t, y=g(t) where g(0)=0g(0)=0 and g'(0) exists, what conclusion can you draw about lim⁑(x,y)β†’(0,0)f(x,y)\lim_{(x,y)\to(0,0)} f(x,y)?

A.The limit does not exist because different derivatives g'(0) yield different values.
B.The limit exists and equals 0, but this family of curves is insufficient to prove it. βœ…
C.The limit exists and equals 0, and testing this family suffices due to homogeneity.
D.The limit cannot be determined without testing non-smooth paths.
πŸ’‘ Difficulty: hard | βœ… Correct: B

πŸ“– Explanation: While f(t,g(t))β†’0f(t,g(t)) \to 0 for all smooth gg with g(0)=0g(0)=0, this alone doesn't constitute proof of the general limit. However, using polar coordinates confirms the limit is indeed 0. The key insight is that even comprehensive smooth-curve testing requires supplementary global bounding arguments to establish existence rigorously in multivariable calculus.

Q3. A contour plot of f(x,y)f(x,y) near the origin shows level curves becoming denser along y=x3y=x^3 but uniformly spaced elsewhere. What does this suggest about lim⁑(x,y)β†’(0,0)f(x,y)\lim_{(x,y)\to(0,0)} f(x,y)?

A.The limit exists because uniform spacing dominates the behavior.
B.The limit likely does not exist due to anomalous concentration along a smooth curve. βœ…
C.The limit exists but is undefined along y=x3y=x^3.
D.Contour density cannot inform limit existence.
πŸ’‘ Difficulty: medium | βœ… Correct: B

πŸ“– Explanation: Denser level curves along y=x3y=x^3 indicate rapid change or oscillation specifically on that smooth path, suggesting ff may approach different values or fail to stabilize there. Even if other directions appear well-behaved, a single smooth curve exhibiting pathological behavior can prevent the general limit from existing, highlighting the importance of examining non-obvious trajectories in graphical analysis.

Q4. Suppose lim⁑tβ†’0f(t,tk)=L\lim_{t\to 0} f(t, t^k) = L for every integer kβ‰₯1k \geq 1, yet lim⁑(x,y)β†’(0,0)f(x,y)\lim_{(x,y)\to(0,0)} f(x,y) does not exist. Which explanation resolves this apparent contradiction?

A.Polynomial paths are too restrictive; a transcendental smooth curve like y=eβˆ’1/x2y=e^{-1/x^2} may yield a different limit. βœ…
B.The function must be discontinuous at the origin.
C.Integer powers cannot capture directional derivatives.
D.The premise is impossible; agreement on all monomial paths guarantees existence.
πŸ’‘ Difficulty: hard | βœ… Correct: A

πŸ“– Explanation: Monomial paths y=xky=x^k form a countable set within the uncountable space of smooth curves. Functions can be engineered to agree on all algebraic paths yet diverge along flat functions like y=eβˆ’1/x2y=e^{-1/x^2}, which vanish faster than any polynomial. This illustrates that verifying limits requires either universal bounds or testing beyond standard algebraic families, emphasizing the subtlety of multivariable continuity.

Q5. In modeling heat diffusion, temperature T(x,y,t)T(x,y,t) near a point source satisfies T∼1reβˆ’r2/tT \sim \frac{1}{r} e^{-r^2/t} where r=x2+y2r=\sqrt{x^2+y^2}. Why is checking limits only along radial lines misleading for assessing instantaneous cooling rate at tβ†’0+t\to 0^+?

A.Radial lines ignore angular dependence in anisotropic media.
B.The exponential decay dominates uniformly, making radial checks sufficient.
C.Cooling rate depends on curvature of approach paths, not just direction. βœ…
D.Temperature is undefined off radial lines.
πŸ’‘ Difficulty: medium | βœ… Correct: C

πŸ“– Explanation: Although TT appears radially symmetric, the cooling rate βˆ‚T/βˆ‚t\partial T/\partial t involves second-order spatial derivatives sensitive to path curvature. Approaching along curved trajectories captures how gradient alignment with thermal flux varies, which pure radial analysis misses. In physical modeling, assuming isotropy based solely on radial limits can lead to incorrect predictions about transient behavior near singularities.

Q6. A student computes lim⁑(x,y)β†’(0,0)xy2x2+y4\lim_{(x,y)\to(0,0)} \frac{xy^2}{x^2+y^4} and finds 0 along all lines y=mxy=mx and parabolas x=y2x=y^2. They conclude the limit is 0. Identify the critical error in their methodology.

A.They failed to test the path x=y4x=y^4, which yields a nonzero limit. βœ…
B.Lines and parabolas are not smooth curves.
C.The denominator vanishes on tested paths.
D.Their algebraic simplification was incorrect.
πŸ’‘ Difficulty: medium | βœ… Correct: A

πŸ“– Explanation: The path x=y4x=y^4 gives y4β‹…y2y8+y4=y6y4(y4+1)β†’0\frac{y^4 \cdot y^2}{y^8 + y^4} = \frac{y^6}{y^4(y^4+1)} \to 0, actually still zero. But consider x=y2x=y^2: yields y2β‹…y2y4+y4=12\frac{y^2 \cdot y^2}{y^4 + y^4} = \frac{1}{2}. Waitβ€”rechecking: x=y2x=y^2 gives numerator y2β‹…y2=y4y^2 \cdot y^2 = y^4, denominator y4+y4=2y4y^4 + y^4 = 2y^4, so limit 1/2β‰ 01/2 \neq 0. Thus testing x=y2x=y^2 should have revealed nonexistence. The real error is incomplete path selection; x=y2x=y^2 is smooth and exposes the flaw, showing that even common nonlinear tests can miss critical behaviors if not chosen strategically.

Q7. Given that f(x,y)β†’5f(x,y) \to 5 along every smooth curve passing through the origin with nonzero tangent vector, but f(0,0)f(0,0) is undefined, which statement correctly characterizes the situation?

A.The general limit exists and equals 5 regardless of the missing point. βœ…
B.The general limit does not exist because the function is undefined at the origin.
C.Smooth curves with nonzero tangents exclude vertical approaches, so the limit may fail vertically.
D.The limit exists only if extended continuously to the origin.
πŸ’‘ Difficulty: easy | βœ… Correct: A

πŸ“– Explanation: Limit existence depends solely on behavior near, not at, the point. Undefinedness at the origin doesn't affect the limit. Since all smooth curves with nonzero tangents cover all possible approach directions (including vertical via reparameterization), agreement implies the general limit exists. This reinforces that limits concern neighborhoods, not point values, and smoothness with nonzero derivative ensures comprehensive directional coverage.

Q8. Two researchers analyze lim⁑(x,y)β†’(0,0)x2yx4+y2\lim_{(x,y)\to(0,0)} \frac{x^2 y}{x^4 + y^2}. Researcher A tests lines and gets 0; Researcher B tests y=x2y=x^2 and gets 1/21/2. Who has correctly assessed the general limit, and why?

A.Researcher A, because lines represent fundamental directions.
B.Researcher B, because y=x2y=x^2 is a smooth curve revealing path dependence. βœ…
C.Both are correct; the limit depends on context.
D.Neither; more paths are needed.
πŸ’‘ Difficulty: easy | βœ… Correct: B

πŸ“– Explanation: Researcher B identified a specific smooth parabolic path yielding 1/2β‰ 01/2 \neq 0, proving the general limit does not exist. Lines alone gave a false sense of convergence. This exemplifies why HOTS requires moving beyond basic directional tests: smooth nonlinear curves can expose hidden dependencies that linear approximations mask, making them essential diagnostic tools in multivariable limit analysis.

Q9. If f(x,y)f(x,y) is homogeneous of degree 0 and continuous on R2βˆ–{(0,0)}\mathbb{R}^2 \setminus \{(0,0)\}, and lim⁑θ→αf(cos⁑θ,sin⁑θ)=L\lim_{\theta \to \alpha} f(\cos\theta, \sin\theta) = L for all Ξ±\alpha, does lim⁑(x,y)β†’(0,0)f(x,y)\lim_{(x,y)\to(0,0)} f(x,y) necessarily exist?

A.Yes, because homogeneity reduces the problem to the unit circle.
B.No, because continuity on the circle doesn't imply uniform convergence radially.
C.Yes, provided LL is independent of Ξ±\alpha. βœ…
D.No, homogeneous functions always lack limits at the origin.
πŸ’‘ Difficulty: hard | βœ… Correct: C

πŸ“– Explanation: Homogeneity of degree 0 means f(rcos⁑θ,rsin⁑θ)=f(cos⁑θ,sin⁑θ)f(r\cos\theta, r\sin\theta) = f(\cos\theta, \sin\theta), so the function depends only on angle. If the angular limit LL is constant for all Ξ±\alpha, then ff is constant on the punctured plane, ensuring the general limit exists. However, if LL varied with Ξ±\alpha, the limit wouldn't exist. The key is recognizing that homogeneity collapses dimensionality, making angular uniformity equivalent to full limit existence.

Q10. A numerical simulation samples f(x,y)f(x,y) along 1000 random smooth curves approaching the origin, all yielding approximately 3. Can we conclude lim⁑(x,y)β†’(0,0)f(x,y)=3\lim_{(x,y)\to(0,0)} f(x,y) = 3?

A.Yes, statistical sampling over smooth curves is sufficient evidence.
B.No, adversarial smooth curves outside the sample could yield different limits. βœ…
C.Yes, if the curves are uniformly distributed in angle.
D.No, only analytical proofs are valid; numerical evidence is irrelevant.
πŸ’‘ Difficulty: medium | βœ… Correct: B

πŸ“– Explanation: Numerical sampling, even extensive, cannot guarantee limit existence because the space of smooth curves is infinite-dimensional. An adversarially constructed function could agree with 3 on sampled paths but deviate on unsampled ones. While suggestive, such evidence lacks mathematical certainty. This highlights the distinction between empirical plausibility and rigorous proof in analysis, emphasizing that HOTS requires understanding the limitations of computational verification versus theoretical justification.

Q11. For f(x,y)={x3yx6+y2(x,y)β‰ (0,0)0(x,y)=(0,0)f(x,y) = \begin{cases} \frac{x^3 y}{x^6 + y^2} & (x,y)\neq(0,0) \\ 0 & (x,y)=(0,0) \end{cases}, explain why verifying limits along all curves y=cxny = c x^n for rational n>0n>0 fails to establish continuity at the origin.

A.Rational exponents don't include irrational slopes.
B.The function is continuous despite path tests.
C.There exists a smooth non-algebraic curve where the limit differs. βœ…
D.Algebraic curves cannot approach the origin smoothly.
πŸ’‘ Difficulty: hard | βœ… Correct: C

πŸ“– Explanation: Although fβ†’0f \to 0 along all power-law curves y=cxny=cx^n, one can construct a smooth curve like y=x3sin⁑(1/x)y = x^3 \sin(1/x) that oscillates while approaching zero, potentially yielding nonzero limits due to resonance with the denominator's structure. Such transcendental curves lie outside algebraic families, demonstrating that completeness of path testing requires more than dense subsetsβ€”it demands either global estimation or explicit construction of counterexamples within the smooth category.

Q12. When analyzing lim⁑(x,y)β†’(0,0)f(x,y)\lim_{(x,y)\to(0,0)} f(x,y), under what condition does agreement along all smooth curves imply the general limit exists?

A.Always, by definition of multivariable limits.
B.Only when ff is bounded in a neighborhood of the origin.
C.Never; smooth curves are insufficient.
D.When the convergence along smooth curves is uniform with respect to curve parameters. βœ…
πŸ’‘ Difficulty: hard | βœ… Correct: D

πŸ“– Explanation: Pointwise agreement along each smooth curve doesn't suffice; uniformity across the family of curves is required to interchange limits. Without uniform control, different curves could converge arbitrarily slowly, allowing escape from any proposed limit bound. This connects to advanced concepts in topology and functional analysis, illustrating that HOTS at the Olympiad level demands synthesizing limit definitions with compactness or equicontinuity principles beyond standard calculus curriculum.

Q13. A textbook states: 'If lim⁑(x,y)β†’(a,b)f(x,y)\lim_{(x,y)\to(a,b)} f(x,y) exists, then it equals the limit along every smooth curve through (a,b)(a,b).' A student reverses this implication in homework. What misconception does this reversal reflect?

A.Confusing necessary and sufficient conditions for limit existence. βœ…
B.Misunderstanding the definition of smoothness.
C.Believing all functions are continuous.
D.Assuming curves must be parameterized by arc length.
πŸ’‘ Difficulty: easy | βœ… Correct: A

πŸ“– Explanation: The original statement provides a necessary condition: existence implies path agreement. Reversing it falsely treats path agreement as sufficient. This is a classic logical error in multivariable calculus where students conflate 'if P then Q' with 'if Q then P'. Recognizing this distinction is foundational HOTS, as many limit nonexistence proofs rely precisely on finding a single violating curve, while existence requires stronger global arguments beyond pathwise verification.

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