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📝 Limits and Continuity in Partial Derivatives (14 MCQs)

📖 From Calculus • 14. Partial Derivatives Calculus • 14 questions available

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What is Limits and Continuity in Partial Derivatives?

Definition:
This topic establishes foundational limit concepts $\lim_{(x,y)\to(a,b)} f(x,y) = L$ required before defining partial derivatives and continuity rigorously.

Example:
Verifying that $\lim_{(x,y)\to(0,0)} \frac{x^2y}{x^2+y^2} = 0$ using polar coordinates before computing $f_x(0,0)$ .

Reason:
Partial derivatives rely on single-variable limits, but multivariable differentiability requires stronger conditions rooted in general multivariable limit theory.

Easy

Medium

Hard

📝 All Limits and Continuity in Partial Derivatives MCQs

Q1. A function $f(x,y)$ is defined such that along every straight line path $y=mx$ passing through the origin, the limit equals 0. However, along the parabolic path $y=x^2$ , the limit equals 1. Which statement best describes the continuity of $f$ at the origin?

A.The function is continuous because all linear paths agree.

B.The function is discontinuous because the limit does not exist due to path dependence. ✅

C.The function is continuous because parabolic paths are negligible in multivariable calculus.

D.The function's continuity cannot be determined without checking polar coordinates.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: This question tests conceptual understanding of multivariable limits. Agreement along infinite straight lines is insufficient for continuity; the limit must be unique regardless of the approach path. Since the parabolic path yields a different value than linear paths, the overall limit does not exist, making the function discontinuous at that point despite linear agreement.

Q2. Consider $f(x,y) = \frac{x^2 y}{x^4 + y^2}$ for $(x,y) \neq (0,0)$ and $f(0,0)=0$ . A student claims the function is continuous at the origin because converting to polar coordinates gives $\lim_{r \to 0} \frac{r^3 \cos^2\theta \sin\theta}{r^4 \cos^4\theta + r^2 \sin^2\theta} = 0$ . Identify the flaw in this reasoning.

A.Polar conversion always proves continuity if the result is zero.

B.The student failed to simplify the trigonometric terms correctly.

C.The denominator can approach zero faster than the numerator for specific

\theta(r)

dependencies, invalidating uniform convergence. ✅

D.The function is actually continuous and the student’s reasoning is flawless.

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: This error analysis question highlights a common misconception: treating $\theta$ as independent of $r$ in polar limits. If $\theta$ varies with $r$ , the expression may not uniformly tend to zero. True continuity requires the limit to hold for all possible approaches, including curved paths where angular dependence on radius creates non-zero limits.

Q3. An engineering model uses $T(x,y) = \frac{\sin(x^2+y^2)}{x^2+y^2}$ to represent temperature distribution near a heat source at the origin. To ensure physical realism, the temperature must be continuous at the source. What value should be assigned to $T(0,0)$ ?

A.0

B.1 ✅

C.Undefined

D.Infinity

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: This application-based direct recall question connects mathematical limits to physical modeling. Recognizing the standard single-variable limit form embedded in two variables, we substitute $u = x^2+y^2$ . As $(x,y) \to (0,0)$ , $u \to 0$ , and $\lim_{u \to 0} \frac{\sin u}{u} = 1$ . Assigning this value ensures continuity and physical consistency.

Q4. Given the contour plot of a function $f(x,y)$ where level curves become infinitely dense and oscillate between values 2 and -2 as they approach the point $(1,1)$ , what can be definitively concluded about $\lim_{(x,y)\to(1,1)} f(x,y)$ ?

A.The limit exists and equals 0 due to symmetry.

B.The limit exists but cannot be read from contours.

C.The limit does not exist because the function fails to settle on a single value. ✅

D.The limit equals 2 since it is the maximum observed value.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: This graph-based interpretation question requires analyzing visual density of level curves. Infinitely dense oscillating contours indicate rapid value changes without convergence. For a limit to exist, function values must stabilize within any epsilon neighborhood. Persistent oscillation between distinct values violates the uniqueness requirement, confirming non-existence regardless of apparent symmetry or boundedness.

Q5. Let $f(x,y) = \begin{cases} \frac{x^3+y^3}{x^2+y^2} & (x,y)\neq(0,0) \\ 0 & (x,y)=(0,0) \end{cases}$ . Without computing partial derivatives, determine whether $f$ is continuous at the origin using inequality bounding.

A.Discontinuous because numerator degree exceeds denominator.

B.Continuous because

|f(x,y)| \leq |x|+|y|

by triangle inequality and squeeze theorem. ✅

C.Continuous because both numerator and denominator are homogeneous polynomials.

D.Discontinuous because the function is undefined along

y=-x

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: This multi-step application problem requires constructing bounds rather than path testing. Using $|x^3| \leq |x|(x^2+y^2)$ and similar for $y^3$ , we get $|f| \leq |x|+|y|$ . Since $|x|+|y| \to 0$ , the squeeze theorem guarantees continuity. This demonstrates how algebraic manipulation can resolve continuity when direct substitution fails.

Q6. A student evaluates $\lim_{(x,y)\to(0,0)} \frac{xy}{\sqrt{x^2+y^2}}$ by claiming it equals 0 because 'the denominator grows while numerator shrinks.' While the answer is correct, why is this reasoning insufficient for full credit?

A.The student should have used L'Hôpital's rule for multivariable functions.

B.The explanation lacks rigorous justification via epsilon-delta or squeeze theorem. ✅

C.The limit actually does not exist and the student guessed correctly.

D.Growth rate comparisons are never valid in multivariable calculus.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: This error analysis question distinguishes intuitive reasoning from mathematical rigor. While growth heuristics often predict correct answers, they fail to address directional dependencies or hidden singularities. Proper justification requires formal tools like squeeze theorem or epsilon-delta proofs that account for all possible approaches, ensuring the conclusion holds universally rather than just intuitively.

Q7. Compare two methods for evaluating $\lim_{(x,y)\to(0,0)} \frac{x^2-y^2}{x^2+y^2}$ : Method A tests paths $y=mx$ ; Method B uses polar coordinates. Which statement accurately assesses their effectiveness?

A.Both methods prove the limit equals 0.

B.Method A shows path dependence proving non-existence; Method B confirms non-uniformity in

\theta

. ✅

C.Method B is superior because polar coordinates always give definitive answers.

D.Method A is invalid because only curved paths matter.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: This mixed concepts question evaluates methodological understanding. Method A reveals different limits for different slopes ( $\frac{1-m^2}{1+m^2}$ ), proving non-existence. Method B yields $\cos(2\theta)$ , showing $\theta$ -dependence confirming the same conclusion. Neither proves existence here, but together they provide complementary evidence of discontinuity through different analytical lenses.

Q8. Define $g(x,y) = \frac{x^a y^b}{x^c + y^d}$ for positive integers $a,b,c,d$ . Under what general condition is $g$ guaranteed continuous at the origin when defined as 0 there?

A.Always continuous regardless of exponents.

B.When

\frac{a}{c} + \frac{b}{d} > 1

using weighted AM-GM inequality. ✅

C.When

a+b > c+d

by simple degree comparison.

D.Only when

a=c

and

b=d

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: This challenging Olympiad-style problem requires synthesizing exponent analysis with inequality techniques. Simple degree comparison fails for mixed denominators. The weighted AM-GM approach shows $x^c + y^d \geq k|x|^{c\alpha}|y|^{d(1-\alpha)}$ . Matching exponents leads to the condition $\frac{a}{c}+\frac{b}{d}>1$ , which ensures the numerator vanishes faster than denominator along all critical paths.

Q9. In a fluid dynamics simulation, velocity field $\vec{V}(x,y) = \left( \frac{-y}{x^2+y^2}, \frac{x}{x^2+y^2} \right)$ models vortex flow. Why can't we define $\vec{V}(0,0)$ to make the field continuous at the origin?

A.Because each component has a removable discontinuity.

B.Because approaching along different angles yields different vector directions, preventing a unique limit. ✅

C.Because the magnitude tends to infinity, violating boundedness.

D.Because partial derivatives don't exist at the origin.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: This scenario-based conceptual question links continuity to physical vector fields. Even though magnitude blows up, the deeper issue is directional non-uniqueness: radial approach gives undefined direction while angular approach preserves rotation. No single vector assignment can satisfy limit definition since neighboring points have fundamentally different orientations, making continuity impossible regardless of magnitude behavior.

Q10. Evaluate $\lim_{(x,y)\to(0,0)} (x^2+y^2)^{x^2+y^2}$ . Which transformation simplifies this indeterminate form most effectively?

A.Convert to Cartesian and apply L'Hôpital twice.

B.Substitute

u=x^2+y^2

reducing to single-variable limit

\lim_{u\to 0^+} u^u

. ✅

C.Use polar coordinates and integrate over theta.

D.Apply logarithmic differentiation directly in two variables.

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: This application question recognizes embedded single-variable structure within multivariable context. The expression depends solely on radial distance $r^2 = x^2+y^2$ , making it radially symmetric. Substituting $u=r^2$ transforms it to $\lim_{u\to 0^+} u^u = e^{\lim u\ln u} = e^0 = 1$ . This avoids unnecessary multivariable complexity by exploiting symmetry.

Q11. A peer argues that since $f(x,y)=\frac{x^2y}{x^2+y^2}$ has partial derivatives $f_x(0,0)=0$ and $f_y(0,0)=0$ , the function must be continuous at the origin. What is the fundamental error?

A.Partial derivatives existing implies differentiability, not necessarily continuity.

B.Continuity requires the full limit to exist, which partial derivatives alone cannot guarantee. ✅

C.The peer miscalculated the partial derivatives.

D.Functions with zero partials are always constant.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: This error analysis targets confusion between partial existence and continuity. Partial derivatives only measure behavior along coordinate axes, ignoring diagonal or curved approaches. A function can have well-defined axial rates of change yet exhibit wild behavior elsewhere. Continuity demands global limit existence, requiring analysis beyond axis-restricted information that partials provide.

Q12. Consider $h(x,y) = \begin{cases} 1 & \text{if } y=x^2 \text{ and } x\neq 0 \\ 0 & \text{otherwise} \end{cases}$ . Despite being zero almost everywhere, why is $h$ discontinuous at the origin?

A.Because the parabola has measure zero.

B.Because every neighborhood of the origin contains points on the parabola where

h=1

, violating epsilon-delta for

\epsilon<1

. ✅

C.Because the function is not differentiable.

D.Because limits along straight lines equal zero.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: This conceptual understanding question challenges intuition about 'almost everywhere' properties. Topological continuity cares about pointwise neighborhoods, not measure. Any delta-ball around origin intersects the parabola, producing function value 1 arbitrarily close to origin where value is 0. This prevents satisfaction of continuity definition regardless of how sparse the exceptional set appears globally.

Q13. When analyzing $\lim_{(x,y)\to(0,0)} \frac{\sin(xy)}{x}$ , a student writes $\approx \frac{xy}{x} = y \to 0$ . While correct, what hidden assumption makes this dangerous in general?

A.Small-angle approximation assumes

xy

is measured in radians.

B.Replacing

\sin(u)

with

u

ignores higher-order terms that may dominate when denominator also vanishes. ✅

C.The cancellation of

x

assumes

x \neq 0

, which is fine for limits.

D.There is no danger; asymptotic equivalence always preserves limits.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: This mixed concepts question examines validity of asymptotic substitutions. While $\sin(u) \sim u$ works here because remainder $O(u^3)/x = O(x^2 y^3) \to 0$ , in cases like $\frac{\sin(x)-x}{x^3}$ , naive replacement loses essential cubic term. Safe usage requires verifying that neglected terms vanish relative to denominator, demanding careful order analysis.

Q14. Suppose $f(x,y)$ satisfies $|f(x,y)| \leq \sqrt{|x|} + |y|^3$ near the origin. Without knowing the explicit formula, what can be concluded about continuity at $(0,0)$ if $f(0,0)=0$ ?

A.Insufficient information without explicit expression.

B.Continuous by squeeze theorem since bound tends to 0. ✅

C.Discontinuous because square root is not differentiable.

D.Continuous only if partial derivatives exist.

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: This direct recall application reinforces squeeze theorem utility with abstract bounds. The given inequality provides dominating function $g(x,y)=\sqrt{|x|}+|y|^3$ that clearly approaches 0 as $(x,y)\to(0,0)$ . Since absolute value of f is trapped between 0 and g, f must also approach 0, matching defined value and establishing continuity purely through bounding.

🔗 Related Topics (MCQs)

📝 Functions of Two or More Variables MCQs 📝 Multivariable function notation and terminology MCQs 📝 Functions of two variables from tables MCQs 📝 Level curves and contour plots MCQs 📝 Graphing functions of two variables MCQs

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