Evaluate the limits in Exercises 33–40 if they exist

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

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

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

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

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

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

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

Thus,

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

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 $3^2+x^2+3xand3+x$is a two-variable polynomial function, it is continuous at all points on $R^2$.

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

At the points where $3+x=0thatisx=-3$, the rational function $\frac{(x^2+3^2+3x)}{(x+3)}$is discontinuous.

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

As a result of the above assertion,

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