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#1 2006-03-03 08:44:47

Chemist
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Limit question

In studying the limits of multi-variable functions, is it reliable to use y=mx and y=x^2 to evaluate the limit of f(x,y) ? Why do we use only these two anyway? And what does a "level curve" mean?

P.S. I know we can use polar coordinates, which is easier.


"Fundamentally one will never be able to renounce abstraction."

Werner Heisenberg

#2 2006-03-03 14:03:56

Ricky
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Re: Limit question

Yes and no.  By using these two, you are usually going to be showing that the limit does not exist.  So you want to end up with is two limits not being equal.

Traveling in abnormal ways is a good thing, such as x^2 and mx.  However, most functions have different limits when approached from the axis.  And besides, normally the axis is easier.

I normally use (x, 0) and (x, x).  These are sufficient for most non-continuous functions, and they are normally easy to plug in.

Last edited by Ricky (2006-03-03 14:04:12)


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

#3 2006-03-03 21:23:53

krassi_holmz
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Re: Limit question

Can't you use y=x+k too?


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#4 2006-03-03 21:26:18

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Re: Limit question

Thanks for your explanation, I think I have a better understanding now. But suppose I'm finding the limit of a function f(x,y) and I substitue y=mx and end up with the limit = m . Can I stop  here and conclude that the limit does not exist simply because it's a variable ?


"Fundamentally one will never be able to renounce abstraction."

Werner Heisenberg

#5 2006-03-03 21:38:45

krassi_holmz
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Re: Limit question

Can you give some example, please?


IPBLE:  Increasing Performance By Lowering Expectations.

#6 2006-03-03 23:17:15

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Re: Limit question

Sure thing krassi_holmz,

f(x,y) = Siny / x

It's required to find the limit of this function as (x,y) ---> (0,y)

So if I take y=mx, lim f(x,y) = Sin(mx) / x = m since Sin(mx) / mx = 1 ( maclaurin series )

Here I think this function has no limit since m is a parameter which belongs to R, it can take any value.

Most often, the question is to find whether a function has a limit. In this case, we're using y=x^2 as well. What we tend to pove is that the function has no limit at all.

f(x,y) = yx^2 / ( x^4 - 2yx^2 + 3y^2 ) if (x,y) ≠ (0,0)
f(x,y) = 0 if (x,y) = (0,0)

Study the limit of this function as ( x,y) ---> (0,0) using y=mx and y=x^2 where m is a real parameter. Show that f doesn't have a limit at the origin O.

for y =mx, the limit of f(x,y) = 0
for y=x^2, the limit of f(x,y) = 1/2

We deduce that the limit is not unique, hence, there's no limit.


"Fundamentally one will never be able to renounce abstraction."

Werner Heisenberg

#7 2006-03-03 23:27:33

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Re: Limit question

Ricky, but notice that if I use y=0 for f(x,y) = yx^2 / ( x^4 - 2yx^2 + 3y^2 ) , the limit will be 0.

If y=x, the limit will be lim x--->0 (x / ( x^2 - 2x + 3 ) ) = 0

What can I conclude from this?


"Fundamentally one will never be able to renounce abstraction."

Werner Heisenberg

#8 2006-03-04 07:02:13

krassi_holmz
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Re: Limit question

Here's a plot of this function:
(this plot isn't correct at x=0 coz' we get division error)


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View Image: sin.gif      

Last edited by krassi_holmz (2006-03-04 07:08:35)


IPBLE:  Increasing Performance By Lowering Expectations.

#9 2006-03-04 07:11:59

krassi_holmz
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Re: Limit question

Sorry for the colors.
I asked Mathematica to compute the limits for different y-s:
Table[Limit[Sin[y]/x,x->0],{y,-10,10}]
And here's what I got(this is table with limits between y=-10 and y=10. The null is when y=0):
{∞, -∞, -∞, -∞, ∞, ∞, ∞, -∞, -∞, -∞, 0, ∞, ∞, ∞, -∞, -∞, -∞, ∞, ∞, ∞, -∞}
Interesting.


IPBLE:  Increasing Performance By Lowering Expectations.

#10 2006-03-04 10:56:47

Ricky
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Re: Limit question

Chemist, nothing is going to work for all functions.  But (x,0) and (x,x) work for a whole lot, and they are easy.

If you plug in (x, mx) and get m, that is enought to say the function is not continuous.


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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