Evaluate the limits in Exercises 33–40 if they exist

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

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.

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

role="math" localid="1653818636152" $\underset{(x,y)\to (x_0,y_0)}{\lim }f(x,y)=f(x_0,y_0)$

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 $x=-2andy=1$ 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,

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