How to prove this function is not continuous at the origin?

Aruwin

@aruwin-WQ7eeu • Oct 24, 2024

Oct 24, 2024

1.2K

Somebody, help me with this question.

Show that the following function is not continuous at the origin(0,0).

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circularsquare

@circularsquare-CAswn0 • Feb 11, 2012

Approach (0,0) along the curve x = y^2 . Then limit of f(x,y) = y^4/2y^4 where y tends to zero.
The limit comes out to be 1/2. Thus , lim f(x,y) along the curve x = y^2 is half at (0,0) whereas f(x,y) = 0 at (0,0) . Since the limit for a given path doesn't match the value at (0,0) , the function isn't continuous at (0,0).
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Aruwin

@aruwin-WQ7eeu • Feb 11, 2012

circularsquare
Approach (0,0) along the curve x = y^2 . Then limit of f(x,y) = y^4/2y^4 where y tends to zero.
The limit comes out to be 1/2. Thus , lim f(x,y) along the curve x = y^2 is half at (0,0) whereas f(x,y) = 0 at (0,0) . Since the limit for a given path doesn't match the value at (0,0) , the function isn't continuous at (0,0).

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Aruwin

@aruwin-WQ7eeu • Feb 11, 2012

circularsquare
Approach (0,0) along the curve x = y^2 . Then limit of f(x,y) = y^4/2y^4 where y tends to zero.
The limit comes out to be 1/2. Thus , lim f(x,y) along the curve x = y^2 is half at (0,0) whereas f(x,y) = 0 at (0,0) . Since the limit for a given path doesn't match the value at (0,0) , the function isn't continuous at (0,0).
Sorry, but can you show me in more detail? Because I don't know how to show the calculation since I don't really understand this.
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circularsquare

@circularsquare-CAswn0 • Feb 11, 2012

read this#-Link-Snipped-#

It contains this specific example as well. It is not solved but hint is given.

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Aruwin

@aruwin-WQ7eeu • Feb 11, 2012

circularsquare
read this#-Link-Snipped-#

It contains this specific example as well. It is not solved but hint is given.
Actually, I get what you mean but I don't know how to find the limits. I have to use l'hopital's rule,right? Can you teach me how to find the limit?

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circularsquare

@circularsquare-CAswn0 • Feb 11, 2012

In case of function of single variable , limit at a point can be found from two directions only i.e from left and right. Since the one variable x is along a line i.e x axis , you can approach the point x = 0 on it only from two directions.
Here only two limits exist right hand limit and left hand limit.

In case of function of two variables the two variables x and y occupy not single line but entire 2-d plane. Thus a point (0,0) can be approached from any path on the 2-d plane. To prove a function is continuous you have to prove that limit along any path you take is constant. To do so you can't actually calculate every limit you have to use general principles. That is not needed here.
To prove a function is 'not' continuous you just have to show any given two limits are not the same.

I asked you to take x = y^2 as one path. Along this path x = y^2. Thus this limit is calculated by substituting x = y^2.
So the limit is y^4/(y^4+ y^4) where y tends to zero.

Of course you can use L'Hospital to calculate. But why not divide by y^4 in numerator and denominator.
See along the path x = y^2 :- f(x,y) = f(y^2 , y) = y^4/2y^4 = 1/2 (because y^4 gets cancelled) . The limit of constant as you approach origin is constant.

Then you can find limit along path y = 0 , it will come out to be zero. Because you have to substitute y = 0 in f(x,y).
Or you could simply say that limit along x = y^2 is 1/2 whereas f(0,0) is zero according to function definition. Hence function is discontinuous.

I hope the concepts are clear now.
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Aruwin

@aruwin-WQ7eeu • Feb 11, 2012

Is

circularsquare
In case of function of single variable , limit at a point can be found from two directions only i.e from left and right. Since the one variable x is along a line i.e x axis , you can approach the point x = 0 on it only from two directions.
Here only two limits exist right hand limit and left hand limit.

In case of function of two variables the two variables x and y occupy not single line but entire 2-d plane. Thus a point (0,0) can be approached from any path on the 2-d plane. To prove a function is continuous you have to prove that limit along any path you take is constant. To do so you can't actually calculate every limit you have to use general principles. That is not needed here.
To prove a function is 'not' continuous you just have to show any given two limits are not the same.

I asked you to take x = y^2 as one path. Along this path x = y^2. Thus this limit is calculated by substituting x = y^2.
So the limit is y^4/(y^4+ y^4) where y tends to zero.

Of course you can use L'Hospital to calculate. But why not divide by y^4 in numerator and denominator.
See along this path x = y^2 , f(x,y) = 1/2 (because y^4 gets cancelled) . The limit of constant as you approach origin is constant.

Then you can find limit along path y = 0 , it will come out to be zero. Because you have to substitute y = 0 in f(x,y).
Or you could simply say that limit along x = y^2 is 1/2 whereas f(0,0) is zero according to functio definition. Hence function is discontinuous.
Opps,sorry. I meant, is it enough by showing 2 limits or is it better if I show the limits of more than 2 functions?
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circularsquare

@circularsquare-CAswn0 • Feb 11, 2012

I think it would do. Because f(0,0) is also given as 0. For continuity, limits along all paths should be equal to each other and also to f(0,0) :- f(0,0) because we have to test continuity at (0,0). It would be easier thus to show it unequal to f(0,0).
Of course you could find limit along path y = 0 too . It comes out to be zero.
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Aruwin

@aruwin-WQ7eeu • Feb 11, 2012

Ah,now I totally get it!!It's not that hard,actually. Thanks,man. Your explanations were really helpful. Now do you mind helping me with my other threads?
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circularsquare

@circularsquare-CAswn0 • Feb 11, 2012

Aruwin
Ah,now I totally get it!!It's not that hard,actually. Thanks,man. Your explanations were really helpful. Now do you mind helping me with my other threads?
I saw the threads , found them tougher than this one. I have forgotten most of the stuff 😉 . This one was simpler , hence attempted .
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Aruwin

@aruwin-WQ7eeu • Feb 11, 2012

circularsquare
I saw the threads , found them tougher than this one. I have forgotten most of the stuff 😉 . This one was simpler , hence attempted .
The other 2 threads are about proving functions,too. I hope you can give them a try. Can you help,please???
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