Q. 39 - Calculus

Question

Evaluate the limits in Exercises 33–40 if they exist 

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

Step-by-Step Solution

Answer

The limit does not exist.

Step 1: Given Information

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

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

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

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

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

Step 2: Defining the limit

Because the value of $\underset{\underset{y=x}{(x,y)\to (3,3)}}{\lim }\frac{x^3-y^3}{x^2-y^2}$ is undetermined, simplify the equation $\frac{x^3-y^3}{x^2-y^2}$.

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

Thus, the expression $\frac{(x-y)(x^2+y^2+xy)}{(x-y)(x+y)}$ is identical to $\frac{(x^2+y^2+xy)}{(x+y)}$ when $y\ne x$,

but the expression $\frac{(x-y)(x^2+y^2+xy)}{(x-y)(x+y)}$ is not equivalent to $\frac{(x^2+y^2+xy)}{(x+y)}$ when $x=y$.

Because the function $\frac{x^3-y^3}{x^2-y^2}$ must be defined on an open set including the point $(3,3)$ in order for the $\underset{\underset{y=x}{(x,y)\to (3,3)}}{\lim }\frac{x^3-y^3}{x^2-y^2}$ to exist.

Because $\frac{x^3-y^3}{x^2-y^2}$ is not defined on the line $y=x$, the function $\frac{x^3-y^3}{x^2-y^2}$ is not defined for an infinitely many points in every open set including the point $(3,3)$.

Thus, the $\underset{\underset{y=x}{(x,y)\to (3,3)}}{\lim }\frac{x^3-y^3}{x^2-y^2}$ does not exist.