Evaluate the limits in Exercises 33–40 if they exist

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

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

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

The function $\underset{\underset{x=3}{(x,y)\to (3,3)}}{\lim }\frac{x^3-y^3}{x^2-y^2}$is

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

$\left\{\frac{0}{0}form\right\}$

Hence, the value of the function $\underset{\underset{}{\left(y\right)\to \left(3\right)}}{\lim }\frac{3^3-y^3}{3^2-y^2}$is in indeterminate form, therefore we change the expression $\frac{3^3-y^3}{3^2-y^2}$.

Thus,

$$\begin{gathered} \underset{\underset{x=3}{(x,y)\to (3,3)}}{\lim }\frac{x^3-y^3}{x^2-y^2}=\underset{\underset{}{\left(y\right)\to \left(3\right)}}{\lim }\frac{3^3-y^3}{3^2-y^2}=\underset{\underset{}{\left(y\right)\to \left(3\right)}}{\lim }\frac{(3-y)(3^2+y^2+3y)}{(3-y)(3+y)} \\ \underset{\underset{}{\left(y\right)\to \left(3\right)}}{\lim }\frac{3^2+y^2+3y}{3+y} \end{gathered}$$

Take the following assertion :

Take a two-variable function $f(x,y)$continuous at the point $(x_0,y_0)\in R^2$.

So, the limit of the function $f(x,y)as(x,y)\to (x_0,y_0)$is defined as

$\underset{\underset{}{(x,y)\to (x_0,y_0)}}{\lim }f(x,y)=f(x_0,y_0)$

Because $3^2+y^2+3yand3+y$is a two-variable polynomial function, it is continuous at all points on $R^2$.

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

At the points where $3+y=0thatis,y=-3$, the rational function $\frac{3^2+y^2+3y}{3^{}+y^{}}$is discontinuous.

Because $y=3$ does not fulfil the equation $y=-3$, the rational function $\frac{3^2+y^2+3y}{3^{}+y^{}}$ is continuous at$y+3$.

As a result of the above assertion,

$$\begin{gathered} \underset{\underset{x=3}{(x,y)\to (3,3)}}{\lim }\frac{x^3-y^3}{x^2-y^2}=\underset{\underset{}{\left(y\right)\to \left(3\right)}}{\lim }\frac{3^2+y^2+3y}{3+y}=\frac{3^2+3^2+3\left(3\right)}{3+3} \\ \frac{9+9+9}{6}=\frac{27}{6}=\frac{9}{2} \end{gathered}$$