Limits & Continuity: Multivariable - Calculus 3

1. Let $w(x,y) = \frac{x^3 + y^3}{x + y}$ for $x + y \neq 0$. To determine if this function can be extended to be continuous on all of $\mathbb{R}^2$, what is the most significant obstacle?
   Answer: The function can be simplified to $x^2 - xy + y^2$ everywhere, so it extends continuously to all points When encountering a rational function with a problematic denominator, your first instinct should be to check if the discontinuity is removable by simplifying the expression algebraically. Let's factor the numerator $x^3 + y^3$ using the sum of cubes formula: $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$. This gives us: $w(x,y) = \frac{x^3 + y^3}{x + y} = \frac{(x + y)(x^2 - xy + y^2)}{x + y}$ For all points where $x + y \neq 0$, we can cancel the common factor: $w(x,y) = x^2 - xy + y^2$ This simplified form is a polynomial, which is continuous everywhere on $\mathbb{R}^2$. Therefore, we can extend the original function by defining $w(x,y) = x^2 - xy + y^2$ for all points, including those where $x + y = 0$. This makes option D correct. Option A is wrong because once we simplify, there are no discontinuities anywhere along $x + y = 0$. Option B incorrectly suggests the extension value is ambiguous—the algebraic simplification gives us a unique, well-defined value at every point. Option C misunderstands the situation entirely; the limit along $x + y = 0$ is perfectly well-defined using the simplified form. Study tip: When you see rational functions with polynomial numerators and denominators, always check if common factors can be canceled. Factoring formulas like sum/difference of cubes often reveal removable discontinuities that make functions extendable to be continuous everywhere.
2. Define $Q(x,y) = \begin{cases} \frac{x^2 - y^2}{x^2 + y^2} \cos\left(\frac{1}{x^2 + y^2}\right) & \text{if } (x,y) \neq (0,0) \\ 0 & \text{if } (x,y) = (0,0) \end{cases}$. What can be concluded about the continuity of $Q$ at the origin?
   Answer: $Q$ is continuous at $(0,0)$ because $\left|\frac{x^2 - y^2}{x^2 + y^2}\right| \leq 1$ and the cosine factor is bounded, so $|Q(x,y)| \to 0$ For continuity at (0,0), we need $\lim_{(x,y) \to (0,0)} Q(x,y) = Q(0,0) = 0$. We have $|Q(x,y)| = \left|\frac{x^2 - y^2}{x^2 + y^2}\right| \cdot \left|\cos\left(\frac{1}{x^2 + y^2}\right)\right| \leq 1 \cdot 1 = 1$. More importantly, as $(x,y) \to (0,0)$, we can write $x^2 + y^2 = r^2$ in polar coordinates, giving $\left|\frac{x^2 - y^2}{x^2 + y^2}\right| = |\cos(2\theta)| \leq 1$. Since the cosine term is bounded and $\frac{x^2 - y^2}{x^2 + y^2} \to \cos(2\theta)$ (which approaches specific values but the overall product approaches 0), the function is continuous. The key is that even though the cosine oscillates, it's bounded, and the limit exists.
3. Consider the function $f(x,y) = \frac{xy^2}{x^2 + y^4}$ for $(x,y) \neq (0,0)$. Which statement best describes the behavior of $f(x,y)$ as $(x,y) \to (0,0)$?
   Answer: The limit does not exist because approaching along $y = x$ gives a different value than along $y = x^{1/2}$ To check if the limit exists, we test different paths. Along $y = x$: $f(x,x) = \frac{x \cdot x^2}{x^2 + x^4} = \frac{x^3}{x^2(1 + x^2)} = \frac{x}{1 + x^2} \to 0$. Along $y = x^{1/2}$: $f(x,x^{1/2}) = \frac{x \cdot x}{x^2 + x^2} = \frac{x^2}{2x^2} = \frac{1}{2}$. Since these paths give different limits, the limit does not exist. Choice A is wrong because degree comparison doesn't apply here. Choice C is incorrect because not all paths give the same value. Choice D confuses the existence of the limit with continuity.
4. Consider the piecewise function $h(x,y) = \begin{cases} \frac{\sin(x^2 + y^2)}{x^2 + y^2} & \text{if } x^2 + y^2 \neq 0 \\ k & \text{if } x^2 + y^2 = 0 \end{cases}$. For what value of $k$ is $h$ continuous at every point in its domain?
   Answer: $k = 1$ because $\lim_{t \to 0} \frac{\sin(t)}{t} = 1$, so continuity requires $h(0,0) = 1$ For continuity at (0,0), we need $\lim_{(x,y) \to (0,0)} h(x,y) = h(0,0) = k$. As $(x,y) \to (0,0)$, we have $x^2 + y^2 \to 0$. Using the standard limit $\lim_{t \to 0} \frac{\sin(t)}{t} = 1$, we get $\lim_{(x,y) \to (0,0)} \frac{\sin(x^2 + y^2)}{x^2 + y^2} = 1$. Therefore $k = 1$. Choice A incorrectly recalls the sine limit. Choice C misunderstands the problem by focusing on maximum values rather than limits. Choice D is wrong because this is exactly a removable discontinuity that can be fixed.
5. The function $T(x,y) = \begin{cases} \frac{x^4 + y^4}{(x^2 + y^2)^2} & \text{if } (x,y) \neq (0,0) \\ c & \text{if } (x,y) = (0,0) \end{cases}$ is given. For which value of $c$ is $T$ continuous everywhere, and what is the key insight needed?
   Answer: $c = \frac{1}{2}$ because $x^4 + y^4 \geq \frac{1}{2}(x^2 + y^2)^2$ and the ratio approaches this lower bound The key insight is to use the inequality $x^4 + y^4 \geq \frac{1}{2}(x^2 + y^2)^2$, which comes from $(x^2 - y^2)^2 \geq 0 \Rightarrow x^4 + y^4 \geq 2x^2y^2 \Rightarrow 2(x^4 + y^4) \geq x^4 + 2x^2y^2 + y^4 = (x^2 + y^2)^2$. Therefore $T(x,y) = \frac{x^4 + y^4}{(x^2 + y^2)^2} \geq \frac{1}{2}$. Along $x = y$: $T(x,x) = \frac{2x^4}{4x^4} = \frac{1}{2}$. Since the function approaches $\frac{1}{2}$ along this path and is bounded below by $\frac{1}{2}$, the limit is $\frac{1}{2}$. Choice A has the wrong value and reasoning. Choice B incorrectly analyzes degrees. Choice D incorrectly claims different path limits.
6. Consider $p(x,y) = \frac{|xy|}{\sqrt{x^2 + y^2}}$ for $(x,y) \neq (0,0)$. Which statement correctly describes the behavior near the origin?
   Answer: $\lim_{(x,y) \to (0,0)} p(x,y) = 0$ because $|xy| \leq \frac{1}{2}(x^2 + y^2)$ provides a squeeze theorem bound Using the inequality $|xy| \leq \frac{1}{2}(x^2 + y^2)$ (from AM-GM), we get $p(x,y) = \frac{|xy|}{\sqrt{x^2 + y^2}} \leq \frac{\frac{1}{2}(x^2 + y^2)}{\sqrt{x^2 + y^2}} = \frac{1}{2}\sqrt{x^2 + y^2}$. As $(x,y) \to (0,0)$, the right side approaches 0, so by squeeze theorem, $\lim_{(x,y) \to (0,0)} p(x,y) = 0$. Choice B is wrong because both $y = x$ and $y = -x$ give the same limit due to the absolute value. Choice C incorrectly computes a specific value. Choice D misunderstands the effect of absolute value on continuity.
7. Let $R(x,y) = \frac{xy^3}{x^2 + y^6}$ for $(x,y) \neq (0,0)$. When investigating $\lim_{(x,y) \to (0,0)} R(x,y)$, which approach most effectively determines whether the limit exists?
   Answer: Test the specific paths $y = x^{1/3}$ and $y = x$ to find conflicting limit values The key insight is that this function requires testing paths where the powers in the denominator balance differently. Along $y = x$: $R(x,x) = \frac{x \cdot x^3}{x^2 + x^6} = \frac{x^4}{x^2(1 + x^4)} = \frac{x^2}{1 + x^4} \to 0$. Along $y = x^{1/3}$: $R(x,x^{1/3}) = \frac{x \cdot (x^{1/3})^3}{x^2 + (x^{1/3})^6} = \frac{x \cdot x}{x^2 + x^2} = \frac{x^2}{2x^2} = \frac{1}{2}$. These give different limits, so the limit doesn't exist. Choice A fails because linear paths all give limit 0. Choice B doesn't reveal the issue clearly. Choice D doesn't help determine if the limit exists.
8. Consider the function $S(x,y) = \frac{\sin(xy)}{x^2 + y^2}$ defined for $(x,y) \neq (0,0)$. If we want to extend $S$ to be continuous at the origin, what value should $S(0,0)$ be assigned?
   Answer: $S(0,0) = 0$ because $|\sin(xy)| \leq |xy|$ and $\frac{|xy|}{x^2 + y^2} \to 0$ as $(x,y) \to (0,0)$ When you encounter a function with a removable discontinuity at the origin, you need to find the limit as $(x,y) \to (0,0)$ to determine the value that makes the function continuous there. For $S(x,y) = \frac{\sin(xy)}{x^2 + y^2}$, the key insight is using the squeeze theorem. Since $|\sin(xy)| \leq |xy|$ for all real values, we have: $|S(x,y)| = \left|\frac{\sin(xy)}{x^2 + y^2}\right| \leq \frac{|xy|}{x^2 + y^2}$ Now, by the Cauchy-Schwarz inequality, $|xy| \leq \frac{x^2 + y^2}{2}$, so: $\frac{|xy|}{x^2 + y^2} \leq \frac{1}{2}$ More directly, as $(x,y) \to (0,0)$, we have $xy \to 0$, so $\frac{|xy|}{x^2 + y^2} \to 0$. By the squeeze theorem, $\lim_{(x,y) \to (0,0)} S(x,y) = 0$. Choice A is incorrect because the average value of sine over neighborhoods is actually 0, not $\frac{1}{2}$. Choice B misapplies the standard limit $\lim_{t \to 0} \frac{\sin(t)}{t} = 1$—this would only work if the denominator were $xy$, not $x^2 + y^2$. Choice C is wrong because the limit actually exists and equals 0 along all paths, as our squeeze theorem argument shows. Study tip: For functions with removable discontinuities at the origin, try the squeeze theorem with trigonometric inequalities like $|\sin(u)| \leq |u|$. This technique works when the "troublesome" part of the numerator is bounded by something that clearly approaches zero.

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