Chapter 12: Q. 33 (page 931)

Evaluate the limits in Exercises 33–40 if they exist.
$\lim_{(x, y) \to (- 2, π)} x^{2} y^{3} \sin y$

Short Answer

Expert verified

The limit is $0$ .

Step by step solution

Given Information

Consider the phrase $\lim_{(x, y) \to (- 2, π)} x^{2} y^{3} \sin y$

Defining the limit

The goal is to assess $\lim_{(x, y) \to (- 2, π)} x^{2} y^{3} \sin y$ if it exists.

Consider the following assertion: Consider a two-variable function $f (x, y)$ that is continuous at all points on $R^{2}$ . The limit of the function $f (x, y)$ as $(x, y) \to (x_{0}, y_{0})$ thus defined as $\lim_{(x, y) \to (x_{0}, y_{0})} f (x, y) = f (x_{0}, y_{0})$

Evaluating the limit

Because $x^{2} y^{3}$ is a two-variable polynomial function, it is continuous for every point on $R^{2}$ , and the transcendental number $\sin y$ is also continuous for every point on $R^{2}$ .

As a result of the statement,

$\lim_{(x, y) \to (- 2, π)} x^{2} y^{3} \sin y = {(- 2)}^{2} {(π)}^{3} \sin π = 0$

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Evaluate the limits in Exercises 33–40 if they exist.
$\lim_{(x, y) \to (- 2, π)} x^{2} y^{3} \sin y$

Short Answer

Step by step solution

Given Information

Defining the limit

Evaluating the limit

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Evaluate the limits in Exercises 33–40 if they exist.lim(x,y)→(-2,π)x2y3siny

Short Answer

Step by step solution

Given Information

Defining the limit

Evaluating the limit

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Discrete Mathematics

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Study anywhere. Anytime. Across all devices.

Evaluate the limits in Exercises 33–40 if they exist.
$\lim_{(x, y) \to (- 2, π)} x^{2} y^{3} \sin y$