Is the Hybrid Function f(x,y) Continuous at the Origin?

In summary, the function is continuous at (0,0) if and only if its limit, as (x, y) goes to (0, 0) is f(0,0).
  • #1
PenKnight
11
0

Homework Statement



f(x,y) is a hybrid function

1. ( ( xy+x )^3 ) / ( x^2 + xy ) when (x,y) not equal to (0,0)
2. 0.........when (x,y) = (0,0)

Is this function continuous at the origin?

Homework Equations



...lim...df |(x,y) = df(0,0)
(x,y)->(0,0)

The Attempt at a Solution



To determine this I have to use the continuity Rule (above) and first principle.

so

lim...f(0,h) - F(0)...(Imagine the divisor line there)
h->0...h.....(and excuse the dots, spaces don't work)

=
lim...[x(y+x^2) - 0 ] x [ 1 ]
h->0..[...x(x+y)...]...[ h ]
...^

From here I'm a little stuck.
Do I cancel the x and solve the limit which result in infinity and therefore the function is not continuous at the origin OR put in the value of (0,h) for (x,y) and get the limit to equal zero?

ie.
...lim...[0(h+0) - 0 ] x [ 1 ]
...h->0..[...0(0+h)...]...[ h ]

..=lim...1
...h->0 ...h
 
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  • #2
I'm extremely confused. You say the problem is to determine whether or not the function is continuous at the origin and refer to the "continuity rule (above)" but I see no "continuity rule". what you have written:
...lim...df |(x,y) = df(0,0)
(x,y)->(0,0)
has little to do with continuity.

You then seem to be using the definition of derivative to try to find the partial derivative with respect to y at (0,0)- I see no reason to want to do that and, if you really want to find it, I don't see why you have not set (x, y) equal to (0, 0) in the calculations from the start.

In any case, a function, f(x,y), is continuous at (0, 0) if and only if its limit, as (x, y) goes to (0, 0) is f(0,0).
I would recommend changing to polar coordinates: [itex]x= r cos(\theta)[/itex], [itex]y= r sin(\theta)[/itex] so [itex](xy+ x)^3= (r^2cos(\theta)sin(theta)+ r cos(theta))^3= r^3cos^3(\theta)(r sin(\theta)+ 1)^3[/itex] and [itex]x^2+ xy= r^2cos^2(\theta)+ r^2sin(\theta)cos(\theta)= r^2cos(\theta)(cos(\theta)+ sin(\theta))[/itex]. Since (x, y) is not (0, 0), r is not 0 so we can cancel the "r"s. The fraction reduces to
[tex]r\frac{cos^3(\theta)(rsin(\theta)+ 1)^3}{cos(\theta)(cos(\theta)+ sin(\theta))}[/tex]
That goes to 0 as r goes to 0, regardless of what [itex]\theta[/itex] is so the function is continuous at (0,0).

If your question was really whether or not the partial differentials are continuous at the origin, that's a different question.
 
  • #3
Yea, I think I got a little confused since my book is a mess.
I had the question and answers messed up.
The continuity Rule should of been f(x) not df and in your exact words
"f(x,y), is continuous at (0, 0) if and only if its limit, as (x, y) goes to (0, 0) is f(0,0)."
I took that from another page in my book to see if the partial derivative was continuous. :S

1.To calculate the partial derivative at (0,0)
2.To show f(x,y) is continuous

1. I think I have to go to first principle to calculate the partial derivative.
2. Use the squeeze theorem to show the limit is zero then it's continuous.
( haven't done much on polar yet but I understand what you have done)

Thanks for your help.
 

1. What is the definition of a limit in a 3d function?

A limit in a 3d function is the value that a function approaches as the input values get closer and closer to a specific point in a three-dimensional space. It can be thought of as the height of the function at a specific point.

2. How is the limit of a 3d function different from a 2d function?

The main difference between the limit of a 3d function and a 2d function is that in a 3d function, the input values are approaching a point in a three-dimensional space, whereas in a 2d function, the input values are approaching a point on a two-dimensional plane.

3. How can you determine the limit of a 3d function graphically?

The limit of a 3d function can be determined graphically by looking at the behavior of the function as the input values approach the specific point. If the function approaches a constant value, then that value is the limit. If the function approaches different values from different directions, then the limit does not exist.

4. What is the relationship between continuity and the limit of a 3d function?

In order for a function to have a limit at a specific point, it must also be continuous at that point. This means that the function has no sudden jumps or breaks at that point. If a function is continuous at a point, then the limit at that point exists and is equal to the function value at that point.

5. Can the limit of a 3d function be calculated algebraically?

Yes, the limit of a 3d function can be calculated algebraically by using the same principles as finding the limit of a 2d function. This involves substituting the specific point into the function and simplifying, taking into account any restrictions or special cases. However, this method can be more complicated and time-consuming compared to graphical or numerical methods.

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