What is true for limit of f (x,y) as (x,y)→(0,1)?

[Lord Popo]

Homework Statement

Let f be a function from R² to R. Suppose that f (x, y) → 3 as (x, y) approaches (0,1) along every line of the form y = kx + 1. What can you say about the limit lim_{(x,y)→(0,1)} f (x, y)? Check the box next to the correct statement.

Homework Equations

N/A

The Attempt at a Solution

*The answer to the problem is "We cannot determine if the limit exists, but if t does, the limit is 3." But I really have no idea why, or what I should attempt to solve this. I'd appreciate any help on this, thank you!

 

[fresh_42]

Imagine a door step of height $3$ (shaped like a wedge) with a vertex along $x=0\,.$

 

[Ray Vickson]

Homework Statement

Let f be a function from R² to R. Suppose that f (x, y) → 3 as (x, y) approaches (0,1) along every line of the form y = kx + 1. What can you say about the limit lim_{(x,y)→(0,1)} f (x, y)? Check the box next to the correct statement.

Homework Equations

N/A

The Attempt at a Solution

*The answer to the problem is "We cannot determine if the limit exists, but if t does, the limit is 3." But I really have no idea why, or what I should attempt to solve this. I'd appreciate any help on this, thank you!

If you take limits along lines y = 1 + kx you are arriving at limits as you come in straight towards the limit point. What about if you approach (0,1) along a spiral, or along some other more complicated curve?

 

[Delta2]

The general guideline for limits like this, is, that if the limit exists it is the same for every particular way that (x,y)->(0,1).In this example you are given that the limit along the lines y=kx+1 exist and it is 3, however this does not guarantee that the limit exists. If we can prove that the limit exists, then as I said before it will be the same for all the possible ways that (x,y)->(0,1) so it will be the same as if (x,y)->(0,1) along the line y=1+kx which gives limit 3.