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Chapter 19: Problem 18

In Exercises 17 through 19, evaluate the given limit by the use of limit theorems. \(\lim _{(x, y) \rightarrow(-2,4)} y \sqrt[3]{x^{3}+2 y}\)

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Step by step solution

01

Understand the Given Limit Expression

Identify the expression whose limit is to be evaluated as \( \lim _{(x, y) \rightarrow(-2,4)} y \sqrt[3]{x^{3}+2 y} \).
02

Substitute Approaching Values

Substitute \(x = -2\) and \(y = 4\) into the expression inside the limit. This results in \[ y \sqrt[3]{x^{3} + 2 y} = 4 \sqrt[3]{(-2)^{3} + 2 \cdot 4} \].
03

Simplify the Inner Expression

Calculate the value inside the cube root: \(-2^3 = -8\), and \(2 \cdot 4 = 8\). So the expression becomes \[ 4 \sqrt[3]{-8 + 8} = 4 \sqrt[3]{0} \].
04

Evaluate the Simplified Expression

The cube root of zero is zero: \( \sqrt[3]{0} = 0 \). So, \(4 \cdot 0 = 0 \).
05

Conclusion

The limit of the given expression as \((x, y)\) approaches \((-2,4)\) is \(0\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Limit Theorems
Limit theorems are useful tools that help us to evaluate limits more easily. They provide rules for how limits of functions behave under various operations. Here are some essential limit theorems that can simplify complex limit calculations:
  • Sum Rule: The limit of a sum of functions is the sum of their limits. If \(\text{lim}_{x \to c} f(x) = L\) and \(\text{lim}_{x \to c} g(x) = M\), then \(\text{lim}_{x \to c} [f(x) + g(x)] = L + M\).
  • Product Rule: The limit of a product of functions is the product of their limits. If \(\text{lim}_{x \to c} f(x) = L\) and \(\text{lim}_{x \to c} g(x) = M\), then \(\text{lim}_{x \to c} [f(x) \times g(x)] = L \times M\).
  • Quotient Rule: The limit of a quotient of functions is the quotient of their limits, provided the limit of the denominator is not zero. If \(\text{lim}_{x \to c} f(x) = L\) and \(\text{lim}_{x \to c} g(x) = M\) with \ M e 0\ , then \ \text{lim}_{x \to c} \frac{f(x)}{g(x)} = \frac{L}{M} \.
  • Power Rule: The limit of a power of a function is the power of the limit of the function. If \(\text{lim}_{x \to c} f(x) = L\), then \(\text{lim}_{x \to c} [f(x)]^n = L^n\).
These rules help to break down complex expressions into more manageable parts, making it easier to find the limit.
Evaluating Limits
Evaluating limits involves finding what value a function approaches as the variables get closer to certain points. Let鈥檚 walk through a related example step-by-step:
Given the limit expression: \ \text{lim}_{(x, y) \rightarrow(-2,4)} y \root{3}{x^{3}+2 y} \
  • Step 1: Identify the expression with the limit to be evaluated: \( y \root{3}{x^{3}+2 y} \)
  • Step 2: Substitute approaching values for \(x\) and \(y\):\(x = -2\) and \(y = 4\). This gives us: \( 4 \root{3}{(-2)^{3} + 2 \times 4} \).
  • Step 3: Simplify the inner expression:\((-2)^{3} = -8 \) and \(2 \times 4 = 8 \), hence: \( 4 \root{3}{-8 + 8} = 4 \root{3}{0} \).
  • Step 4: Evaluate the simplified expression: \( \root{3}{0} = 0 \), thus \( 4 \times 0 = 0 \).
  • Step 5: Conclude the result: The limit of the given expression as \( (x, y) \) approaches \( (-2,4) \) is \( 0 \).
By following this structured approach, we can systematically evaluate the limit and ensure that we haven't missed any step.
Continuity
Continuity is a key concept in calculus that describes whether a function is seamless at a point or over an interval. If a function is continuous at a point, there鈥檚 no sudden jump, hole, or break at that point. To check for continuity at a point \((a, b)\) for a function \(f(x, y)\), three conditions should be satisfied:
  • The function must be defined at the point: \(f(a, b)\) should exist.
  • The limit must exist at that point: \( \text{lim}_{(x, y) \rightarrow (a, b)} f(x, y) \) must exist.
  • The value of the function at the point should equal the limit: \( \text{lim}_{(x, y) \rightarrow (a, b)} f(x, y) = f(a, b) \).
Let's consider the function and limit from our previous example: \( y \root{3}{x^{3}+2 y} \). To check for continuity at \((-2,4)\):
  • 1. The function exists at \((-2,4)\).
  • 2. The limit exists and we found it to be 0.
  • 3. The value of the function at \((-2,4)\) matches our limit: \(4 \times 0 = 0 \).
Hence, \( y \root{3}{x^{3}+2 y} \) is continuous at \((-2,4)\). Understanding continuity helps in describing the behavior of functions and their stability over specified points.

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Most popular questions from this chapter

Water is flowing into a tank in the form of a right-circular cylinder at the rate of \(\frac{4}{5} \pi \mathrm{ft}^{3} / \mathrm{min}\). The tank is stretching in such a way that even though it remains cylindrical, its radius is increasing at the rate of \(0.002 \mathrm{ft} / \mathrm{min}\). How fast is the surface of the water rising when the radius is \(2 \mathrm{ft}\) and the volume of water in the tank is \(20 \pi \mathrm{ft}^{3}\) ?

In Exercises 8 through 17, determine the region of continuity of \(f\) and draw a sketch showing as a shaded region in \(R^{2}\) the region of continuity of \(f\). $$ f(x, y)= \begin{cases}\frac{x^{2}-y^{2}}{x-y} & \text { if } x \neq y \\ x-y & \text { if } x=y\end{cases} $$

In Exercises 11 through 14, find the total derivative \(d u / d t\) by two methods: (a) Use the chain rule; (b) make the substitutions for \(x\) and \(y\) or for \(x, y\), and \(z\) before differentiating. $$ u=y e^{x}+x e^{y} ; x=\cos t ; y=\sin t $$

Given \(f(x, y)= \begin{cases}\frac{3 x^{2} y}{x^{2}+y^{2}} & \text { if }(x, y) \neq(0,0) \\ 0 & \text { if }(x, y)=(0,0)\end{cases}\) This function is continuous at \((0,0)\) (see Example 3, Sec. 19.2, and Illustration 1, Sec. 19.3). Prove that \(D_{1} f(0,0)\) and \(D_{2} f(0,0)\) exist but \(D_{1} f\) and \(D_{2} f\) are not continuous at \((0,0)\).

In Exercises 1 through 8 , do each of the following: (a) Find \(D_{11} f(x, y)\); (b) find \(D_{22} f(x, y) ;\) (c) show that \(D_{12} f(x, y)=D_{21} f(x, y) .\) $$ f(x, y)=x \cos y-y e^{x} $$
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