91Ó°ÊÓ

The goal is to discover$f(x,y)andg(x,y)$functions, as well as a location $(a,b)∈R^2$such that

$\left(\underset{\underset{}{(x,y)→(a,b)}}{\lim }f(x,y)\right)\left(\underset{\underset{}{(x,y)→(a,b)}}{\lim }g(x,y)\right)≠\underset{\underset{}{(x,y)→(a,b)}}{\lim }\left(f\right(x,y\left)g\right(x,y\left)\right).$

The goal is to find a function whose limit is known at a given location and which can be divided into two halves.

Consider the functions$f\left(x\right)=\frac{x+y}{x-y}andg\left(x\right)=\frac{x-y}{x+y}$, respectively.

Consider the point$(a,b)=(0,0)$.

These are of the form $\frac{0}{0}$at the given location for the individual functions. As a result, neither limit exists at this time. The limit of the product of two functions is $\underset{\underset{}{(x,y)→(0,0)}}{\lim }\left(\frac{x+y}{x-y}\right)\left(\frac{x-y}{x+y}\right)=\underset{\underset{}{(x,y)→(0,0)}}{\lim }\left(1\right)=1$

As a result, the product of individual function limits is not equal to the product of individual function limits.

It appears to contradict the general rule of product of limits, which stipulates that

$\left(\underset{\underset{}{(x,y)→(a,b)}}{\lim }f(x,y)\right)\left(\underset{\underset{}{(x,y)→(a,b)}}{\lim }g(x,y)\right)=\underset{\underset{}{(x,y)→(a,b)}}{\lim }\left(f\right(x,y\left)g\right(x,y\left)\right).$

The existence of both individual function limitations is required to implement this rule. As a result, this rule only applies if the limit of functions $f(x,y)andg(x,y)$at point $(a,b)∈R^2$does exists.

However, the preceding example disregards this constraint. As a result, the example cannot be regarded to violate the limit product rule.