# Continuity of piecewise function f(x,y)

## Homework Statement

Determine all points at which the given function is continuous.

For practice, I want to verify the continuity. Moreover, with piecewise function, I have to verify continuity anyway.

Q1 Q2 ## The Attempt at a Solution

Let's do the second problem first.
For Q2, we have x =/= y for the first condition, whereas if x = y, we let f(x,y) = 2x.
So to check continuity, we would have to make sure
(1) f(a) is defined,
(2) limit as x goes to a exists, and
(3) f(a) = limit as x goes to a

The piecewise function proved that f(a) is defined for x = y, then we have 2x (for an arbitrary number, possible).

Then I did limit.
limit as y goes to (x = y), we get x^2 - y^2/ x - y, after simplification, we had (x +y) remained. Substituted y = x, we have x+x, so the limit is 2x.

Indeed, f(x=y) = 2x = limit as y goes to x=y
This problem required no squeeze theorem because simplification and substitution worked.

For the first problem, I attempted a few ways.
We know #1 is true, when x^2+y = 1, we have to have f(x,y) = 1.
So I tried to find its limit and see if it would come out to be 1.

So I took limit of the first function. I can't simplify it, so I am down with squeeze theorem and path methods. But I was confused what limit to take. I tried to eliminate one of the parameter, and did L'hospital rule. This allowed me to take the derivative of sin^1/2(...) which would put its derivative with a -1/2 power. Thus I had

-2y (sqrt(1-x^2-y^2)) / -2y cos(1-x^2-y^2) , and cancel -2y.
But I am stuck. If I did x^2 + y^2 = 1 and solve for x (or y), and plug it in, I would always get -x^2 or -y^2 inside the square root, which is not right...

Please tell me whether my approach for Q2 was right, and how to solve for Q1. Thank you.

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It helps to know that $\lim_{x\to 0}\frac{sin(x)}{x}= 1$

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## Homework Statement

Determine all points at which the given function is continuous.

For practice, I want to verify the continuity. Moreover, with piecewise function, I have to verify continuity anyway.

Q1 Q2 ## The Attempt at a Solution

Let's do the second problem first.
For Q2, we have x =/= y for the first condition, whereas if x = y, we let f(x,y) = 2x.
So to check continuity, we would have to make sure
(1) f(a) is defined,
(2) limit as x goes to a exists, and
(3) f(a) = limit as x goes to a

The piecewise function proved that f(a) is defined for x = y, then we have 2x (for an arbitrary number, possible).

Then I did limit.
limit as y goes to (x = y), we get x^2 - y^2/ x - y, after simplification, we had (x +y) remained. Substituted y = x, we have x+x, so the limit is 2x.

Indeed, f(x=y) = 2x = limit as y goes to x=y
This problem required no squeeze theorem because simplification and substitution worked.
You have the basic idea, but you're being sloppy in your explanation. For example,
• f(a) - the function has two variables, not one, so f(a) is meaningless in this problem.
• limit as y goes to (x = y) - I get what you're trying to say, but you're not saying it very well.
The function z = f(x, y) is defined in a piecewise fashion, with one definition for points along the line x = y, and the other for all other points in the plane. Your limit should start off looking like this:
$$\lim_{(x, y) \to (a, b)} f(x, y)$$

The limit can be written more simply for points on the line y = x.
For the first problem, I attempted a few ways.
We know #1 is true, when x^2+y = 1, we have to have f(x,y) = 1.
Typo here. When x^2 + y^2 = 1, f(x, y) is defined to be equal to 1.
So I tried to find its limit and see if it would come out to be 1.

So I took limit of the first function. I can't simplify it, so I am down with squeeze theorem and path methods. But I was confused what limit to take. I tried to eliminate one of the parameter, and did L'hospital rule. This allowed me to take the derivative of sin^1/2(...) which would put its derivative with a -1/2 power. Thus I had

-2y (sqrt(1-x^2-y^2)) / -2y cos(1-x^2-y^2) , and cancel -2y.
??? The function is sin(sqrt(1 - x^2 - y^2)). Where did the sin go?

Also, you can't just "take the derivative" - this is a function of two variables. You would have to take the partial derivative with respect to either x or y.

Note that your function has two definitions: one for points on the circle x^2 + y^2 + 1, and the other for points inside that circle. That circle can be represented in polar form very simply, and that will help you do what you need to do.
But I am stuck. If I did x^2 + y^2 = 1 and solve for x (or y), and plug it in, I would always get -x^2 or -y^2 inside the square root, which is not right...

Please tell me whether my approach for Q2 was right, and how to solve for Q1. Thank you.