Find functions $f(x,y)andg(x,y)$and a point $(a,b)\in R^2$such that $\underset{\underset{}{(x,y)\to (a,b)}}{\lim }f(x,y)+\underset{\underset{}{(x,y)\to (a,b)}}{\lim }g(x,y)\ne \underset{\underset{}{(x,y)\to (a,b)}}{\lim }\left(f\right(x,y)+g(x,y\left)\right).$Does this example contradict the sum rule for limits of a function of two variables?

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

$\underset{\underset{}{(x,y)\to (a,b)}}{\lim }f(x,y)+\underset{\underset{}{(x,y)\to (a,b)}}{\lim }g(x,y)\ne \underset{\underset{}{(x,y)\to (a,b)}}{\lim }\left(f\right(x,y)+g(x,y\left)\right).$

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 $f\left(x\right)=\frac{x}{x+y}andg\left(x\right)=\frac{y}{x+y}$, 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 sum of two functions is$\underset{\underset{}{(x,y)\to (0,0)}}{\lim }\left(\frac{x}{x+y}+\frac{y}{x+y}\right)=\underset{\underset{}{(x,y)\to (0,0)}}{\lim }\left(\frac{x+y}{x+y}\right)=\underset{\underset{}{(x,y)\to (0,0)}}{\lim }\left(1\right)=1$

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

It appears to contradict the general rule of sum of limits, which stipulates that $\underset{\underset{}{(x,y)\to (a,b)}}{\lim }f(x,y)+\underset{\underset{}{(x,y)\to (a,b)}}{\lim }g(x,y)=\underset{\underset{}{(x,y)\to (a,b)}}{\lim }\left(f\right(x,y)+g(x,y\left)\right).$

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)$$f$at point localid="1653554624584" $(a,b)\in R^2$does exists.

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