Q. 38 - Calculus

Question

Evaluate the limits in Exercises 33–40 if they exist 

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

Step-by-Step Solution

Answer

The limit is $\frac{9}{2}$.

Step 1: Given Information

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

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

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

Step 2: Defining the limit

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

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

Step 3: Evaluating the limit

Because $3^2+x^2+3x and 3+x$ 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 $3+x=0 that is x=-3$, 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,

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