25f166…: “With a humid cube”

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Select all of the following statements below that are true.

If $\Lim {x}{1} F(x,1) = 5$ and $\Lim {y}{2} F(2,y)=5$ , then $\Lim {\vector {x,y}}{\vector {1,2}} F(x,y) =5$ . If $F(x,y)$ is continuous at $(3,5)$ , then $\Lim {\vector {x,y}}{\vector {3,5}} F(x,y)$ exists. If $\Lim {\vector {x,y}}{\vector {3,5}} F(x,y)$ exists, then $F(x,y)$ is continuous at $(3,5)$ . If $F(x,y)$ is not continuous at $(3,5)$ , then $\Lim {\vector {x,y}}{\vector {3,5}} F(x,y)$ does not exist. If $F(x,y)$ approaches the same value along every straight line path through $(a,b)$ , then $\Lim {\vector {x,y}}{\vector {a,b}} F(x,y)$ must exist.

Here are some explanations.

For the first choice, note that the given information tells you the values that the function approaches along the lines $x=1$ and $y=2$ is the same. This is not enough to conclude that the limit exists. As a counterexample, consider the function below.
$F(x,y) = \begin {cases} 5, & x=1 \\ 5, & y=2 \\ 0, & \textrm {otherwise } \end {cases}$
For the second option, the definition of continuity requires not only that $\Lim {\vector {x,y}}{\vector {3,5}} F(x,y)$ exists, but that $F(3,5)$ is defined and that $\Lim {\vector {x,y}}{\vector {3,5}} F(x,y)=F(3,5).$
To be continuous at $(3,5)$ , we need that $\Lim {\vector {x,y}}{\vector {3,5}} F(x,y)$ exists AND $\Lim {\vector {x,y}}{\vector {3,5}} F(x,y) = F(3,5)$ .
The issue for continuity may not be the nonexistence of the limit; it may be because both $\Lim {\vector {x,y}}{\vector {3,5}} F(x,y)$ and $F(3,5)$ exist, but do not line up. For instance, consider the function below.
$F(x,y) = \begin {cases} 0, & (x,y) \neq (3,5) \\ 1, & (x,y) =(3,5) \end {cases}$
We need the function to approach the same value along every path through $(a,b)$ , not just all straight line paths. A counterexample is the function in the text, or the function below, which approaches $0$ along all straight line paths through $(x,y) = (0,0)$ , but approaches $\frac {1}{2}$ as $(x,y) \to (0,0)$ along $y=x^2$ .
$F(x,y) = \begin {cases} \frac {x^2y}{x^4+y^2}, & (x,y) \neq (0,0) \\ 0, & (x,y) =(0,0) \end {cases}$