Limits & Continuity: Multivariable - Multivariable Calculus

1. Consider the function $f(x,y) = \frac{x^2 y}{x^4 + y^2}$. A student calculates the limit of $f(x,y)$ as $(x,y) \to (0,0)$ along any line $y=mx$ and finds that $\lim_{x \to 0} f(x, mx) = 0$ for all real numbers $m$. Based on this information, what can be concluded about $\lim_{(x,y) \to (0,0)} f(x,y)$?
   Answer: The limit does not exist, because the limit value depends on the path of approach to the origin. The fact that the limit is 0 along all straight-line paths $y=mx$ is not sufficient to prove the existence of the overall limit. To show the limit does not exist, we must find two paths that yield different limit values. Along the path $y=x^2$, the function becomes $f(x, x^2) = \frac{x^2(x^2)}{x^4 + (x^2)^2} = \frac{x^4}{2x^4} = \frac{1}{2}$ for $x \neq 0$. The limit along this path is $1/2$. Since the limit along linear paths is 0 and the limit along the parabolic path $y=x^2$ is $1/2$, the overall limit does not exist.
2. Let $h(x,y) = \begin{cases} \frac{x^2y}{x^2 + 2y^2} & \text{if } (x,y) \neq (0,0) \\ 0 & \text{if } (x,y) = (0,0) \end{cases}$. Which statement correctly describes the continuity and differentiability of $h$ at $(0,0)$?
   Answer: $h$ is continuous but not differentiable at $(0,0)$ because the partial derivatives exist but are not continuous there For continuity: $|h(x,y)| = |\frac{x^2y}{x^2 + 2y^2}| \leq |\frac{x^2y}{2y^2}| = \frac{|x|^2}{2}$ when $y \neq 0$. More precisely, $|h(x,y)| \leq \frac{|x|^2|y|}{x^2 + 2y^2} \leq \frac{|x||y|}{\sqrt{2}} \leq \frac{x^2 + y^2}{2\sqrt{2}} \to 0$, so $h$ is continuous. For partial derivatives: $h_x(0,0) = \lim_{t \to 0} \frac{h(t,0) - 0}{t} = 0$ and $h_y(0,0) = \lim_{t \to 0} \frac{h(0,t) - 0}{t} = 0$. However, $h_x(x,y) = \frac{4xy^3}{(x^2 + 2y^2)^2}$ for $(x,y) \neq (0,0)$. Along $y = x$, $h_x(x,x) = \frac{4x^4}{(3x^2)^2} = \frac{4}{9}$, which doesn't approach $h_x(0,0) = 0$. So partial derivatives exist but aren't continuous, meaning $h$ is not differentiable. Choices B and D incorrectly conclude differentiability. Choice C is wrong about continuity.
3. Let $f(x,y) = (x^2+y^2) \cos(\frac{1}{x^2+y^2})$ for $(x,y) \neq (0,0)$. Which statement correctly describes the limit of $f(x,y)$ as $(x,y) \to (0,0)$?
   Answer: The limit is 0, which can be established by applying the Squeeze Theorem with appropriate bounding functions. We can use the Squeeze Theorem to evaluate this limit. We know that for any non-zero argument, $-1 \le \cos(\frac{1}{x^2+y^2}) \le 1$. Multiplying all parts of this inequality by the non-negative term $x^2+y^2$ gives $-(x^2+y^2) \le (x^2+y^2) \cos(\frac{1}{x^2+y^2}) \le (x^2+y^2)$. As $(x,y) \to (0,0)$, both $-(x^2+y^2)$ and $(x^2+y^2)$ approach 0. Therefore, by the Squeeze Theorem, $\lim_{(x,y) \to (0,0)} f(x,y) = 0$.
4. On what set $S$ is the function $f(x,y) = \frac{\ln(x^2+y^2-1)}{x}$ continuous?
   Answer: $S = \{(x,y) \in \mathbb{R}^2 \mid x^2+y^2 > 1 \text{ and } x \neq 0 \}$. A function is continuous on its domain. We need to find the domain of $f(x,y)$. There are two conditions for $f(x,y)$ to be defined. First, the argument of the natural logarithm must be strictly positive: $x^2+y^2-1 > 0$, which means $x^2+y^2 > 1$. This is the set of points outside the unit circle. Second, the denominator cannot be zero, so $x \neq 0$. The function is continuous on the set of points satisfying both conditions simultaneously, which is the region outside the unit circle, excluding the y-axis.
5. Suppose $f: \mathbb{R}^2 \to \mathbb{R}$ is a function for which $\lim_{(x,y) \to (a,b)} f(x,y) = L$. Which of the following statements must be true as a consequence?
   Answer: For any real number $m$, the limit $\lim_{x \to a} f(x, b+m(x-a))$ exists and is equal to $L$. The definition of the limit $\lim_{(x,y) \to (a,b)} f(x,y) = L$ means that $f(x,y)$ approaches $L$ as $(x,y)$ approaches $(a,b)$ along any path. Statement B describes the limit along any straight-line path through $(a,b)$, which is a specific subset of all possible paths. If the general limit exists, the limit along any specific path must also exist and be equal to the general limit. Statement A is the definition of continuity, which is a stronger condition than the existence of a limit. Statement C is the converse, which is false; checking a finite number of paths (or even all linear paths) is not sufficient. Statement D is also not necessarily true; a function can have a limit at a point without being differentiable there.
6. Consider the function $F(x,y) = \begin{cases} \frac{x^3 - y^3}{x^2 + y^2} & \text{if } (x,y) \neq (0,0) \\ L & \text{if } (x,y) = (0,0) \end{cases}$. For what value of $L$ is $F$ continuous at $(0,0)$, and what can be concluded about the partial derivatives $F_x(0,0)$ and $F_y(0,0)$?
   Answer: $L = 0$; $F_x(0,0) = 1$ and $F_y(0,0) = -1$ based on the directional behavior of the numerator For continuity, we need $\lim_{(x,y) \to (0,0)} \frac{x^3 - y^3}{x^2 + y^2} = L$. Using polar coordinates: $\frac{r^3\cos^3\theta - r^3\sin^3\theta}{r^2} = r(\cos^3\theta - \sin^3\theta) \to 0$ as $r \to 0$, so $L = 0$. For partial derivatives: $F_x(0,0) = \lim_{h \to 0} \frac{F(h,0) - F(0,0)}{h} = \lim_{h \to 0} \frac{h^3/h^2 - 0}{h} = \lim_{h \to 0} \frac{h}{h} = 1$. Similarly, $F_y(0,0) = \lim_{h \to 0} \frac{F(0,h) - F(0,0)}{h} = \lim_{h \to 0} \frac{-h^3/h^2 - 0}{h} = \lim_{h \to 0} \frac{-h}{h} = -1$. Choice A incorrectly assumes partial derivatives are 0. Choice C is wrong about continuity. Choice D incorrectly claims partial derivatives don't exist.
7. For the function $f(x,y) = \frac{e^{x^2+y^2} - 1}{x^2+y^2}$, the limit as $(x,y) \to (0,0)$ exists. What value should be assigned to $f(0,0)$ to make the function continuous at the origin?
   Answer: $1$ When you encounter a function that's undefined at a point but need to determine continuity, you're dealing with removable discontinuities. The key is finding the limit as you approach that point and seeing if assigning that limit value makes the function continuous. Here, $f(x,y) = \frac{e^{x^2+y^2} - 1}{x^2+y^2}$ is undefined at $(0,0)$ because both numerator and denominator equal zero. To find $\lim_{(x,y) \to (0,0)} f(x,y)$, let $u = x^2 + y^2$. As $(x,y) \to (0,0)$, we have $u \to 0$, so: $\lim_{(x,y) \to (0,0)} \frac{e^{x^2+y^2} - 1}{x^2+y^2} = \lim_{u \to 0} \frac{e^u - 1}{u}$ This is the standard limit definition of the derivative of $e^u$ at $u = 0$. Since $\frac{d}{du}e^u = e^u$, we get $e^0 = 1$. Therefore, defining $f(0,0) = 1$ makes the function continuous. Choice A ($0$) would create a jump discontinuity since the limit is $1$, not $0$. Choice B is incorrect because we just showed the limit exists and equals $1$, so continuity is achievable by proper definition. Choice C ($e$) confuses the general derivative formula $\frac{d}{dx}e^x = e^x$ with this specific limit where the exponent approaches $0$. Study tip: When you see indeterminate forms like $\frac{0}{0}$ in multivariable limits, look for connections to single-variable derivative definitions, especially with exponential and trigonometric functions. The limit $\lim_{h \to 0} \frac{e^h - 1}{h} = 1$ is fundamental and appears frequently.
8. Let $f(x,y) = x^2 + y^2 - 1$ and $g(t) = \sqrt{t}$. The composite function $h(x,y) = g(f(x,y))$ is continuous on which of the following sets?
   Answer: The set of all points $(x,y)$ in $\mathbb{R}^2$ such that $x^2+y^2 \ge 1$. The function $f(x,y)$ is a polynomial and is continuous everywhere on $\mathbb{R}^2$. The function $g(t) = \sqrt{t}$ is continuous on its domain, which is $t \ge 0$. The composite function $h(x,y) = g(f(x,y))$ is continuous wherever $f(x,y)$ is in the domain of $g$. This requires the argument of the square root to be non-negative, so we must have $f(x,y) \ge 0$. This implies $x^2+y^2-1 \ge 0$, which simplifies to $x^2+y^2 \ge 1$. This set represents all points on or outside the unit circle centered at the origin.
9. Consider the piecewise function $G(x,y) = \begin{cases} \frac{\sin(x^2 + y^2)}{x^2 + y^2} & \text{if } x^2 + y^2 \neq 0 \\ k & \text{if } x^2 + y^2 = 0 \end{cases}$. For $G$ to be continuous everywhere, what must $k$ equal, and what additional property does $G$ have at $(0,0)$?
   Answer: $k = 1$; $G$ is differentiable at $(0,0)$ but $\nabla G$ is not continuous there because $G_x$ and $G_y$ are not continuous at $(0,0)$ For continuity at $(0,0)$: $\lim_{(x,y) \to (0,0)} \frac{\sin(x^2 + y^2)}{x^2 + y^2} = \lim_{t \to 0^+} \frac{\sin t}{t} = 1$, so $k = 1$. For differentiability: $G_x(0,0) = \lim_{h \to 0} \frac{G(h,0) - 1}{h} = \lim_{h \to 0} \frac{\frac{\sin(h^2)}{h^2} - 1}{h} = \lim_{h \to 0} \frac{\sin(h^2) - h^2}{h^3}$. Using $\sin u = u - \frac{u^3}{6} + O(u^5)$: $\sin(h^2) = h^2 - \frac{h^6}{6} + O(h^{10})$, so $G_x(0,0) = \lim_{h \to 0} \frac{-h^6/6}{h^3} = 0$. Similarly $G_y(0,0) = 0$. For $(x,y) \neq (0,0)$: $G_x = \frac{2x\cos(x^2+y^2)(x^2+y^2) - 2x\sin(x^2+y^2)}{(x^2+y^2)^2}$. As $(x,y) \to (0,0)$, this doesn't approach 0, so $\nabla G$ isn't continuous at $(0,0)$. Choices A and C have wrong conclusions about differentiability. Choice B has wrong value for $k$.
10. Consider the function defined by $f(x,y) = \frac{x^2-y^2}{x^2+y^2}$. Which of the following statements provides the most accurate description of the limit of $f(x,y)$ as $(x,y) \to (0,0)$?
    Answer: The limit does not exist because the limit value depends on the angle of approach $\theta$. To analyze this limit, we convert to polar coordinates by setting $x = r\cos\theta$ and $y = r\sin\theta$. The expression becomes $f(r, \theta) = \frac{r^2\cos^2\theta - r^2\sin^2\theta}{r^2\cos^2\theta + r^2\sin^2\theta} = \frac{r^2(\cos^2\theta - \sin^2\theta)}{r^2(\cos^2\theta + \sin^2\theta)} = \cos(2\theta)$. The value of the function depends only on the angle $\theta$ and not on the radius $r$. As we approach the origin $(r \to 0)$, the value of the function can be any value in $[-1, 1]$ depending on the path's angle. For example, along the x-axis ($\theta=0$), the limit is $\cos(0)=1$. Along the y-axis ($\theta=\pi/2$), the limit is $\cos(\pi)=-1$. Since the limit value is path-dependent, the limit does not exist.
11. Let $\Lambda(x,y,z) = \begin{cases} \frac{xyz}{x^2 + y^2 + z^2} & \text{if } (x,y,z) \neq (0,0,0) \\ 0 & \text{if } (x,y,z) = (0,0,0) \end{cases}$. Which statement correctly analyzes the continuity of $\Lambda$ at the origin?
    Answer: $\Lambda$ is continuous at $(0,0,0)$ because in spherical coordinates, $\Lambda(\rho\sin\phi\cos\theta, \rho\sin\phi\sin\theta, \rho\cos\phi) = \rho f(\phi,\theta)$ where $f$ is bounded Using spherical coordinates $x = \rho\sin\phi\cos\theta$, $y = \rho\sin\phi\sin\theta$, $z = \rho\cos\phi$: $\Lambda = \frac{\rho^3\sin^2\phi\cos\phi\cos\theta\sin\theta}{\rho^2} = \rho\sin^2\phi\cos\phi\cos\theta\sin\theta$. Since $|\sin^2\phi\cos\phi\cos\theta\sin\theta| \leq |\sin^2\phi\cos\phi| \leq \frac{2\sqrt{2}}{9}$ (maximum when $\sin\phi = \sqrt{\frac{2}{3}}$), we have $|\Lambda| \leq C\rho$ for some constant $C$, so $\Lambda \to 0$ as $\rho \to 0$. Choice A uses wrong inequality ($|xyz| \leq \frac{(x^2+y^2+z^2)^{3/2}}{3\sqrt{3}}$ by AM-GM, but this doesn't give the stated bound). Choice B: along $(t,t,t)$, $\Lambda = \frac{t^3}{3t^2} = \frac{t}{3} \to 0$. Choice D is incorrect; while different paths give different approach rates, they all approach 0.
12. Let the function $f$ be defined as $f(x,y) = 1$ if $x$ and $y$ are both rational numbers, and $f(x,y) = 0$ otherwise. Which of the following statements best describes the continuity of this function?
    Answer: The function is continuous nowhere on the plane $\mathbb{R}^2$. This function is a two-dimensional analogue of the Dirichlet function. To be continuous at a point $(a,b)$, the limit $\lim_{(x,y) \to (a,b)} f(x,y)$ must exist and equal $f(a,b)$. However, in any arbitrarily small neighborhood (an open disk) around any point $(a,b)$, there exist points where both coordinates are rational and points where at least one coordinate is irrational. This means that in any neighborhood of any point, the function takes on both values, 0 and 1. The function values do not approach a single value $L$ as $(x,y)$ approaches $(a,b)$. Therefore, the limit does not exist at any point in $\mathbb{R}^2$, and the function is continuous nowhere.
13. Consider the function $f(x,y) = \frac{x^2y}{x^4 + y^2}$ for $(x,y) \neq (0,0)$. Which statement best describes the behavior of $f(x,y)$ as $(x,y)$ approaches $(0,0)$?
    Answer: The limit does not exist because approaching along $y = mx^2$ gives different values for different values of $m$ To check if the limit exists, we test different paths. Along $y = mx^2$: $f(x,mx^2) = \frac{x^2(mx^2)}{x^4 + (mx^2)^2} = \frac{mx^4}{x^4 + m^2x^4} = \frac{m}{1 + m^2}$. This gives different values for different $m$ (e.g., $m=1$ gives $\frac{1}{2}$, $m=2$ gives $\frac{2}{5}$), so the limit doesn't exist. Choice A is wrong because degree comparison doesn't work here. Choice C is incorrect because the polar analysis would give $|f| \leq r|\cos\theta|$ which suggests the limit is 0, but this misses the problematic paths. Choice D is wrong because the function being undefined on axes doesn't prevent a limit from existing.
14. Let a function be defined as: $f(x,y) = \begin{cases} 0 & \text{if } y \le 0 \text{ or } y \ge x^2 \\ 1 & \text{if } 0 < y < x^2 \end{cases}$ Which of the following best describes the limit of this function as $(x,y) \to (0,0)$?
    Answer: The limit does not exist, because paths can be chosen to approach the origin from within the region where $f=1$ and from the region where $f=0$. When evaluating limits of piecewise functions in multivariable calculus, you need to check whether the function approaches the same value along all possible paths to the point. If different paths yield different limiting values, the limit does not exist. Let's analyze this function's behavior near the origin. The key insight is understanding the geometry: the parabola $y = x^2$ passes through $(0,0)$, creating a boundary between regions where $f = 1$ (below the parabola, above the x-axis) and $f = 0$ (elsewhere). Now consider different approaches to $(0,0)$. If you approach along the x-axis where $y = 0$, you're always in the region where $f = 0$. However, if you approach along a path like $y = \frac{x^2}{2}$ (which stays below the parabola $y = x^2$ near the origin), you're in the region where $f = 1$. Since these paths give different limiting values, the limit doesn't exist. Answer A correctly identifies this reasoning. Answer B is wrong because the existence of paths where $f = 1$ doesn't establish the limit—you need consistency across all paths. Answer C makes an error: linear paths $y = mx$ with $m > 0$ actually enter the region where $f = 0$ only when $mx \geq x^2$, which happens when $x \geq m$, but near the origin when $x < m$, we have $mx < x^2$, so $f = 1$. Answer D incorrectly suggests limits can be "averages"—limits are about consistency, not weighted values. Study tip: For piecewise function limits, always test multiple path types, especially along boundary curves, not just straight lines.
15. Consider the function $f(x,y) = \frac{2xy}{x^2+y^2}$ for $(x,y) \neq (0,0)$ and $f(0,0)=0$. Which of the following statements accurately describes the function at the origin $(0,0)$?
    Answer: The function is not continuous at $(0,0)$, but its partial derivatives $f_x(0,0)$ and $f_y(0,0)$ both exist. When analyzing continuity and differentiability at a point in multivariable calculus, you need to check these properties independently—one doesn't guarantee the other. Let's first check continuity at $(0,0)$. For continuity, we need $\lim_{(x,y) \to (0,0)} f(x,y) = f(0,0) = 0$. Let's approach along the line $y = mx$: $\lim_{x \to 0} \frac{2x(mx)}{x^2 + (mx)^2} = \lim_{x \to 0} \frac{2mx^2}{x^2(1+m^2)} = \frac{2m}{1+m^2}$ This limit depends on $m$, meaning different paths give different limits. Since the limit doesn't exist, the function is not continuous at $(0,0)$. Now for partial derivatives. Using the definition: $f_x(0,0) = \lim_{h \to 0} \frac{f(h,0) - f(0,0)}{h} = \lim_{h \to 0} \frac{0 - 0}{h} = 0$ Similarly, $f_y(0,0) = 0$. Both partial derivatives exist. Answer D is correct: the function is not continuous but has existing partial derivatives at $(0,0)$. Answer A incorrectly assumes that existing partial derivatives imply continuity—this is false in multivariable calculus. Answer B incorrectly claims the partial derivatives don't exist when they clearly do. Answer C incorrectly states the function is continuous when we showed it's not. Remember: In multivariable calculus, partial derivatives can exist even when the function isn't continuous. Always test continuity using limits along different paths, and compute partial derivatives using the definition when working at potentially problematic points.
16. Define $\phi(x,y) = \begin{cases} \frac{x^4 + y^4}{(x^2 + y^2)^{3/2}} & \text{if } (x,y) \neq (0,0) \\ 0 & \text{if } (x,y) = (0,0) \end{cases}$. What is the correct analysis of $\phi$ near $(0,0)$?
    Answer: $\phi$ is continuous at $(0,0)$ because in polar coordinates, $\phi(r\cos\theta, r\sin\theta) = r(\cos^4\theta + \sin^4\theta) \to 0$ as $r \to 0$ Converting to polar coordinates: $\phi(r\cos\theta, r\sin\theta) = \frac{r^4(\cos^4\theta + \sin^4\theta)}{(r^2)^{3/2}} = \frac{r^4(\cos^4\theta + \sin^4\theta)}{r^3} = r(\cos^4\theta + \sin^4\theta)$. Since $\cos^4\theta + \sin^4\theta$ is bounded (actually between $\frac{1}{2}$ and 1), we have $|\phi(x,y)| \leq r \to 0$ as $r \to 0$. So $\phi$ is continuous at $(0,0)$. Choice A is wrong about the limit value. Choice C: along $x = y$, $\phi(t,t) = \frac{2t^4}{(2t^2)^{3/2}} = \frac{2t^4}{2^{3/2}t^3} = \frac{t}{\sqrt{2}} \to 0$. Choice D gives wrong reasoning (degree comparison doesn't directly apply with fractional exponents).
17. Define $\Xi(x,y) = \begin{cases} \frac{e^{x^2 + y^2} - 1}{x^2 + y^2} & \text{if } (x,y) \neq (0,0) \\ \alpha & \text{if } (x,y) = (0,0) \end{cases}$. For $\Xi$ to be continuous at $(0,0)$, what must $\alpha$ equal, and what can be said about the differentiability of $\Xi$ there?
    Answer: $\alpha = 1$; $\Xi$ is not differentiable at $(0,0)$ because the partial derivatives exist but are not continuous there For continuity: $\lim_{(x,y) \to (0,0)} \frac{e^{x^2 + y^2} - 1}{x^2 + y^2} = \lim_{t \to 0^+} \frac{e^t - 1}{t} = 1$, so $\alpha = 1$. For partial derivatives: $\Xi_x(0,0) = \lim_{h \to 0} \frac{\Xi(h,0) - 1}{h} = \lim_{h \to 0} \frac{\frac{e^{h^2} - 1}{h^2} - 1}{h} = \lim_{h \to 0} \frac{e^{h^2} - 1 - h^2}{h^3}$. Using $e^u = 1 + u + \frac{u^2}{2} + \frac{u^3}{6} + O(u^4)$: $e^{h^2} = 1 + h^2 + \frac{h^4}{2} + O(h^6)$, so $\Xi_x(0,0) = \lim_{h \to 0} \frac{h^4/2}{h^3} = 0$. Similarly $\Xi_y(0,0) = 0$. For $(x,y) \neq (0,0)$: $\Xi_x = \frac{2xe^{x^2+y^2}(x^2+y^2) - 2x(e^{x^2+y^2}-1)}{(x^2+y^2)^2}$. As $(x,y) \to (0,0)$, this approaches $\lim \frac{2x \cdot 1 \cdot 0 - 2x \cdot 0}{0}$, which requires L'Hôpital's rule analysis and doesn't simply approach 0. The partial derivatives are not continuous at $(0,0)$. Choices A and D incorrectly conclude about continuity of $\nabla \Xi$. Choice B has wrong $\alpha$.

Start this quiz