- #1

- 280

- 0

## 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

## Homework Equations

## 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.