Q. 33 - Calculus

Question

Evaluate the limits in Exercises 33–40 if they exist. 

$\underset{(x,y)\to (-2,\pi )}{\lim }{x}^2{y}^3\sin y$

Step-by-Step Solution

Answer

The limit is $0$.

Step 1: Given Information

Consider the phrase $\underset{(x,y)\to (-2,\pi )}{\lim }{x}^2{y}^3\sin y$

Step 2: Defining the limit

The goal is to assess $\underset{(x,y)\to (-2,\pi )}{\lim }{x}^2{y}^3\sin y$ 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 $\underset{(x,y)\to (x_0,y_0)}{\lim }f(x,y)=f(x_0,y_0)$

Step 3: 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,

$\underset{(x,y)\to (-2,\pi )}{\lim }{x}^2{y}^3\sin y = {(-2)}^2{\left(\pi \right)}^3\sin \pi = 0$