Q. 36 - Calculus

Question

Evaluate the limits in Exercises 33–40 if they exist 

$\underset{(x,y)\to (-2,1)}{\lim }\frac{x^3-y^3}{x^2-y^2}$

Step-by-Step Solution

Answer

The limit is $-3$.

Step 1: Given Information

Consider the phrase $\underset{(x,y)\to (-2,1)}{\lim }\frac{x^3-y^3}{x^2-y^2}$

The goal is to assess $\underset{(x,y)\to (-2,1)}{\lim }\frac{x^3-y^3}{x^2-y^2}$ if it exists.

Step 2: Defining the limit

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 $f(x,y) as (x,y)\to (x_0,y_0)$ is then 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 $x^3-y^3 and x^2-y^2$ is a two-variable polynomial function, it is continuous at all points on $R^2$.

As a result, the rational function $\frac{x^3-y^3}{x^2-y^2}$ is continuous at all positions where $\frac{x^3-y^3}{x^2-y^2}$ is defined.

At the places where $\frac{x^3-y^3}{x^2-y^2}$the rational function is discontinuous at $x^2-y^2=0$,that is

$$\begin{gathered} x^2=y^2 \\ x=\pm y \end{gathered}$$

Because $x=-2 and y=1$ do not satisfy the equation $x=\pm y$, the rational function $\frac{x^3-y^3}{x^2-y^2}$ is continuous at $(-2,1).$

As a result of the statement,

$\underset{(x,y)\to (-2,1)}{\lim }\frac{x^3-y^3}{x^2-y^2}=\frac{{(-2)}^3-1}{{(-2)}^2-1}=-\frac{9}{3}=-3$