91Ó°ÊÓ

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

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

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

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

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

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

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

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 $y≠x$,

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