Directional Derivatives and Limits


by autarch
Tags: derivatives, directional, limits
autarch
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#1
Mar31-12, 10:16 AM
P: 9
How can I use the directional derivative of a two variable function to show that the limit does not exist? For example, suppose I have a function f(x,y)=g(x)/f(y) and g(a)=f(b)=0 and the limit as x and y go to a and b is 0. How would I use the directional derivative to show that the limit at (a,b) does not exist.

So far, I have tried to take the directional derivatives of the f(x,y) at points around the (a,b), but I feel this is inconclusive because nothing is known about the function itself, other than the fact that it is undefined at (a,b).
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Woopydalan
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#2
Mar31-12, 12:17 PM
P: 746
you can choose different paths, and if the paths aren't equal, the limit does not exist.
autarch
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#3
Mar31-12, 01:31 PM
P: 9
Quote Quote by Woopydalan View Post
you can choose different paths, and if the paths aren't equal, the limit does not exist.
That does not resolve my issue. If the only information that I have is that g(x)/f(y) is discontinuous at (a,b), a directional derivative along different paths won't show anything due to the ambiguity of the function. Furthermore, if the limit does not exist at (a,b), then I cannot use the gradient at (a,b). At least, I don't think I can.

HallsofIvy
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#4
Mar31-12, 01:53 PM
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Directional Derivatives and Limits


What you have written really does not make much since. You say "f(x,y)=g(x)/f(y)" so that you are using f both for a function of 1 variable and a function of two variables. I think what you intend would be better written "f(x,y)= g(x)/h(y)".

In any case, it is impossible to give an answer to your question without more information as to what f and g are like close to 0.
autarch
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#5
Mar31-12, 02:39 PM
P: 9
Quote Quote by HallsofIvy View Post
What you have written really does not make much since. You say "f(x,y)=g(x)/f(y)" so that you are using f both for a function of 1 variable and a function of two variables. I think what you intend would be better written "f(x,y)= g(x)/h(y)".

In any case, it is impossible to give an answer to your question without more information as to what f and g are like close to 0.
Good point and thank you. Also, I noticed that I have made a mistake. f(x,y) should equal
g(y)/(f(x) not g(x)/f(y). As far as providing more information, g(x)/h(y) represents any function that is undefined at a particular point. The only information that is given is that the line x=a is not in the domain, and the answer can be shown with a directional derivative, although any method is acceptable; however, I must show that the limit of f(x,y) at (a,b) does not exist.

As far as describing what f(x) and g(y) are like close to 0, I do not know, since f(x) and g(y) merely represent one variable functions, but nothing specific.
micromass
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#6
Mar31-12, 04:39 PM
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Try to approach is coming from a direction of the x-axis and of the y-axis. Are these two limits the same?
autarch
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#7
Mar31-12, 05:08 PM
P: 9
Quote Quote by micromass View Post
Try to approach is coming from a direction of the x-axis and of the y-axis. Are these two limits the same?
I don't understand what you mean, but the only thing I am trying to prove is that the limit as (x,y)->(a,b) does not exist for g(y)/h(x). I am trying to do this with directional derivatives, and the line x=a is not in the domain of f(x,y)=g(y)/h(x).
autarch
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#8
Mar31-12, 05:57 PM
P: 9
Could I just simply show that the directional derivative at (a-1,b) in the direction of (a+1,b) is different from the directional derivative at (a-1,b+1) in the direction of (a+1,b+1)?
Cecilia48
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#9
Mar31-12, 08:20 PM
P: 8
That does not resolve my issue.
autarch
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#10
Mar31-12, 08:32 PM
P: 9
Quote Quote by Cecilia48 View Post
That does not resolve my issue.
What issue is that?
autarch
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#11
Apr1-12, 04:25 PM
P: 9
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