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Ricky
2006-03-04 10:56:47

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.

krassi_holmz
2006-03-04 07:11:59

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.

krassi_holmz
2006-03-04 07:02:13

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

Chemist
2006-03-03 23:27:33

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?

Chemist
2006-03-03 23:17:15

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.

krassi_holmz
2006-03-03 21:38:45

Can you give some example, please?

Chemist
2006-03-03 21:26:18

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 ?

krassi_holmz
2006-03-03 21:23:53

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

Ricky
2006-03-03 14:03:56

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.

Chemist
2006-03-03 08:44:47

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.