A limit in two variables

  • #26
Yes, but I had the inequalities going the wrong way before.

For small ##h > 0## we have
$$\ln (1+h) > h^2 - \frac 1 2 h^2$$
Apply this to the numerator ##\ln(1 + 2x^2+y^2) > \ln(1 + x^2+y^2) = \ln(1+r^2),## where ##r^2 = x^2+y^2.##

In the denominator we have ##(x^2 + 3 y^2)^2 < (3 x^2 + 3 y^2)^2 = 9 r^4,##
so the function satisfies
$$f = \frac{\ln(1+2x^2 + y^2)}{(x^2 + 3 y^2)^2} > \frac{r^2 - \frac 1 2 r^4}{9 r^4}= \frac 1 9 \left(\frac{1}{r^2} - \frac{1}{2} \right)$$
thanks man, this should do the trick
Converting to polar form for this function doesn't help, because the polar form results in different values for different values of the angle. Converting to polar form is helpful in cases where ##\theta## is no longer present.
still think it doesnt include linear paths, or the theta would be a variable not a constant, and in that case you couldnt treat it anymore as a one variable limit and you have the same problem as before. btw, theta still appears in the function i was trying to find the limit.
 
  • #27
i've been tought that I need to discover another condition constantly greater than this one that goes to 0 at (0,0) to discover an answer. or on the other hand if the arrangement doesnt exist, attempt to discover two ways with two distinct outcomes in 0,0. whichever way I picked, this goes to +infinite. in class I have never observed a limit in two factors that exists and goes to +inf, having a vagary frame, so I dont have an idea on the best way to understand it. any thoughts?
hope you didnt register to this forum just for this
 
  • #28
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still think it doesnt include linear paths, or the theta would be a variable not a constant
If ##\theta## is no longer present, it can have any value, so isn't constant. You're still thinking that r has to approach 0 along some straight line with ##\theta## constant, and that isn't so.
 
  • #29
Ray Vickson
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thanks man, this should do the trick

still think it doesnt include linear paths, or the theta would be a variable not a constant, and in that case you couldnt treat it anymore as a one variable limit and you have the same problem as before. btw, theta still appears in the function i was trying to find the limit.

I don't understand your claim: the polar method (with fixed ##\theta\:##) does, in fact, include linear paths. That is, when ##\theta## is fixed, the paths ##r \to 0## are straight lines pointing to ##(0,0)##. If a limit fails to exist for at least one value of ##\theta## the 2D-limit fails to exist, and you are done. If the limit exists for each ##\theta## but is different for some different values of ##\theta## then, again, the 2D-limit fails to exist and you are done.

The only problematical case is when the "polar" limit exists and is the same for all values of ##\theta##, because that means that all straight-line limits exist and are independent of the line you choose. However, we know from examples that a 2D-limit can still fail to exist in this case, because going to zero along a curved path may give you different answers. Those types of questions are inherently tricky, and there seems to be no well-known standard methods available for treating them; it is as though each question stands on its own, and it is hard to generalize from one example to another.
 
  • #30
I don't understand your claim: the polar method (with fixed ##\theta\:##) does, in fact, include linear paths. That is, when ##\theta## is fixed, the paths ##r \to 0## are straight lines pointing to ##(0,0)##. If a limit fails to exist for at least one value of ##\theta## the 2D-limit fails to exist, and you are done. If the limit exists for each ##\theta## but is different for some different values of ##\theta## then, again, the 2D-limit fails to exist and you are done.

The only problematical case is when the "polar" limit exists and is the same for all values of ##\theta##, because that means that all straight-line limits exist and are independent of the line you choose. However, we know from examples that a 2D-limit can still fail to exist in this case, because going to zero along a curved path may give you different answers. Those types of questions are inherently tricky, and there seems to be no well-known standard methods available for treating them; it is as though each question stands on its own, and it is hard to generalize from one example to another.
everything you are saying is correct, every single word. but now you are going against mark44, because he says that the polar form includes also paths that are not straight lines.
 
  • #31
Ray Vickson
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everything you are saying is correct, every single word. but now you are going against mark44, because he says that the polar form includes also paths that are not straight lines.

Not really: we are saying different things. When ##r## and ##\theta## are not linked we get straight lines for fixed ##\theta## and varying ##r##. However, when ##r = R(\theta)## or ##\theta =\Theta(r)## we are just looking at the polar equation for some curve, and in that case anything goes. Mark44 was talking about varying both ##r## and ##\theta##, but I was talking about something different.
 
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  • #32
what you are saying is what i have been saying since my first comment. mark is saying that converting to the polar form we are talking about considers also NON linear paths (i never said that the polar form didnt consider linear paths as you implied), which is the opposite as what you have stated. what would even be the point of finding a polar form such as r(theta)?
 
  • #33
it wouldnt even be possible for the function i was trying to solve, and certainly it wouldnt help for finding a limit
 
  • #34
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what you are saying is what i have been saying since my first comment.
No, it is not. You said this in post #4.
i think what you are saying is incorrect, the polar form doesnt include non linear paths, so you cant be sure if the limits exist.

Kenneth1997 said:
mark is saying that converting to the polar form we are talking about considers also NON linear paths
Yes, and both Ray Vickson and I agree on this, if you read closely what he has written. You have at least one misconception that seems to be keeping you from understanding, and that is, if ##\theta## doesn't appear in a polar equation, ##\theta## is necessarily constant. That is NOT true.
Simple example: Consider the equation x = 1 in the Cartesian plane. y doesn't appear in this equation, so it can take on any real value. The graph of x = 1 is a vertical line through the point (1, 0). Every point on this line has an x-coordinate of 1, but the y-coordinate is completely arbitrary.
 
  • #35
epenguin
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Have you tried just two simple paths, a horizontal one and a vertical one? I.e. The limit as x→0 when y = 0. And if it has one then the other way around. Do they exist? And if they do are they the same? Maybe this would not always be conclusive, but seems to me it is here and is not difficult.
 
  • #36
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Thread closed for Moderation...

After a Mentor discussion and some thread cleanup, the thread will remain closed. Thanks to everyone who contributed to the thread. :smile:
 
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