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Proving limits exist

by mathrocks
Tags: exist, limits, proving
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mathrocks
#1
Jun16-04, 08:30 AM
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As of now I'm completely confused on how to prove that a limit of several variables exist. I know that if you plug a bunch of functions in and you get the same results you can start to think the limit exists but without proof it's pointless...
So for example:

lim (x^2 (sin^2 y)) / (x^2+2y^2)
(x,y)->(0,0)

I know it involves using the squeeze theorem but thats about all I know. Any help would be appreciated.
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matt grime
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Jun16-04, 08:44 AM
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for all x,y, with x not zero, x^2/(x^2+2y^2)=1/(1+2(y/x)^2) < 1, thus the original expression is less than sin^2y, when x is zero the quantity is obviously zero (y not also zero), hence...
Gokul43201
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Jun16-04, 09:31 AM
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For a single variable function you approach x0 from both sides (on the real line) and show that both these limits are the same.

For a 2-variable function, I would magine that you'd have to approach (x0,y0) from each of the (infinitely) many directions on the real plane, and arrive at the same limit for them all.

Clearly, this not practically doable...so there must be another approach.

matt grime
#4
Jun16-04, 09:55 AM
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Proving limits exist

the definition is for f(x,y) to tend to a limit L is that given any e>0, there exists a d>0, such that sqrt(x^2+y^2)<d => |f(x,y)-L|<e
mathrocks
#5
Jun16-04, 12:37 PM
P: 106
Quote Quote by matt grime
for all x,y, with x not zero, x^2/(x^2+2y^2)=1/(1+2(y/x)^2) < 1, thus the original expression is less than sin^2y, when x is zero the quantity is obviously zero (y not also zero), hence...

How did you get 1/(1+2(y/x)^2)? Where did y/x come from?
master_coda
#6
Jun16-04, 01:39 PM
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Quote Quote by mathrocks
How did you get 1/(1+2(y/x)^2)? Where did y/x come from?
He divided both the top and the bottom of the fraction by x^2. Thus x^2 becomes 1 and y^2 becomes (y/x)^2.
HallsofIvy
#7
Jun16-04, 09:36 PM
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Generally speaking the simplest way to show that a limit exists is:

1) translate the point to (0,0) (if the point is (a,b), replace x with x'-a and y with y'-b).

2. change into polar coordinates. Since the point of the limit is that we get close to a single value as (x,y) gets close to (0,0), changing to polar coordinates makes that depend on the single variable r.
mathwonk
#8
Jul23-04, 02:29 PM
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I am puzzled by the suggestion of Halls ofIvy that one can eliminate a variable by using polar coordinates. Wouldn't there still be a theta variable for the angle?
mathwonk
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Jul23-04, 02:30 PM
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Let me give an example. If the function is just plain theta, i.e. the angle, then along any radial direction the function is constant hence ahs a limit. But overall there is no one value that the angle is tending to as you approach zero.
arildno
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Jul23-04, 02:42 PM
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Quote Quote by mathwonk
Let me give an example. If the function is just plain theta, i.e. the angle, then along any radial direction the function is constant hence ahs a limit. But overall there is no one value that the angle is tending to as you approach zero.
But this is PRECISELY why it is so smart to introduce polar coordinates!!

A limit exists at a point (x,y) if and only if letting the radius go to 0 yields a unique answer
EDIT:
In particular, if we end up with a non-constant function g(theta), we now that there cannot exist a limit.
Galileo
#11
Jul23-04, 02:54 PM
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I think Matt Grime's solution is the easiest. Just use:

[tex]0 \leq \frac{x^2sin^2y}{x^2+y^2} \leq sin^2y[/tex]

and apply the squeeze theorem.
mathwonk
#12
Jul23-04, 04:56 PM
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Arildno, I believe we are not understanding each other. if you mean to substitute the radius for the delta, in an epsilon delta proof then I agree, but if you mean look at the radial limits and see if they are all the same, then I disagree that this is equivalent to the existence of the limit.

I.e. I concur that if the radial limit g(theta) is a non constant function then there is no limit. But the subtle thing seems to me that even if that function IS constant then there is still no guarantee there IS a limit.

I.e. even if the same limit is reached for all angles, but reached at a different RATE for different angles, then it is possible that given epsilon, there is no distance delta from the origin such that for all points at that distance, the function value is within epsilon of the common radial limit.

I.e. given epsilon, the delta that works radially is a function of theta, and if this function not bounded below, we have a problem.

Does this make sense?

I am not differing with Matt Grimes solution to this problem which seems elegant and correct.
mathwonk
#13
Jul23-04, 05:21 PM
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This is a fun question. I tried to cook up an example doing what I said was possible before. Arildno, please check me on this.

Let f(r,theta) = r^(theta), where r is the radius of the point, and theta ranges over the interval (0,2pi]. Then for any fixed value of theta, as r goes to zero, this goes to zero. Thus the radial limit function g(theta) is constant, always equal to zero, since every radial limit is zero.

On the other hand if we fix r, no matter how small, then for small enough theta, the number r^(theta) is close to 1, since this is a continuous function of theta, and r^0 = 1. Thus no matter how small r is f(r,theta) is never less than a small epsilon for every theta at once.

Thus the limit of f(r,theta) does not exist at the origin, even though all the radial limits exist and are equal.

Does this seem right?
mathwonk
#14
Jul23-04, 05:23 PM
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by the way, courant's calculus book (vol2?) is a great source of examples like the previous one, where say all directional derivatives exist in all directions at the origin, and yet the function is not differentiable at the origin.
Gza
#15
Jul23-04, 07:03 PM
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mathrocks,

I hate to overload you with more techniques and such, since everyone here is already doing a good job of showing you, but what works for me is to make the y variable in f(x,y) a function of x, and simply plug it in for all the y values on the RHS. Make it general by using a k term, such as y=kx, or y = kx^2, or y = kx^3, you get the picture. If after simplifying and taking the limit, a k is still hanging around, you know that the limit depends on value of k, therefore, the limit doesn't exist (this is actually pretty much the same idea posted by HallsofIvy with the polar coordinate thinking) Working enough of these kinds of problems will allow you to see what function of x (or y for that matter) to use, to try to "break" the problem, and show that the limit doesn't exist.
mathwonk
#16
Jul24-04, 09:44 PM
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I think I finally understand what Halls of Ivy and Arildno have been trying to tell me. I.e. write the function f(x,y) as a function of (r,theta), say h(r,theta). If h(r,theta) goes to L as r goes to 0, uniformly in theta, then f has as limit L as (x,y) approaches (0,0).

I.e. if for every epsilon >0, there is a delta >0, such that for all 0< r < delta, and all theta, we have h(r,theta) < epsilon, then f(x,y) also approaches L as (x,y) approaches (0,0).

for example we can try this on the example above f(x,y) =

[x^2 sin^2(y)]/[x^2+y^2], and get h(r,theta) =

r^2 cos^2(theta)sin^2(rsin(theta))/r^2 =

cos^2(theta) sin^2(rsin(theta)). Then as r goes to zero, so does sin^2(rsin(theta)), independently of theta.

Since also cos^2(theta) is bounded, the whole thing goes to zero.

This is no better than Matt Grimes proof, but as Halls of Ivy suggested, this provides a template for doing it without being too clever.

thank you!!
HallsofIvy
#17
Jul25-04, 10:08 AM
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And thank God we don't have to be too clever!!

By the way, if the problem is to find the limit as (x,y) goes to (a,b) then first "translate" the whole problem: replace x with x'+a and y with y'+ b so that, as (x,y) goes to (a,b), (x',y') goes to (0,0) and THEN change to polar coordinates.


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