# A limit in two variables

• Kenneth1997

## Homework Statement

lim (x,y)->(0,0) (ln(1+2x^2+y^2))/(x^2+3y^2)^2

## The Attempt at a Solution

i've been tought that i have to find another equation always bigger than this one that goes to 0 at (0,0) to find a solution. or if the solution doesn't exist, try to find two paths with two different results in 0,0. whichever path i chose, this goes to +infinite. in class i have never seen a limit in two variables that exists and goes to +inf, having an indetermination form, so i don't have a clue on how to solve it. any ideas?

## Homework Statement

lim (x,y)->(0,0) (ln(1+2x^2+y^2))/(x^2+3y^2)^2

## The Attempt at a Solution

[/B]
i've been tought that i have to find another equation always bigger than this one that goes to 0 at (0,0) to find a solution. or if the solution doesn't exist, try to find two paths with two different results in 0,0. whichever path i chose, this goes to +infinite. in class i have never seen a limit in two variables that exists and goes to +inf, having an indetermination form, so i don't have a clue on how to solve it. any ideas?

If you choose a path and it gives + infinity, then the limit can't exist (if it would exist, it would give the limit value along every path).

1. lim (x,y)->(0,0) (ln(1+2x^2+y^2))/(x^2+3y^2)^2

3. I've been tought that i have to find another equation always bigger than this one that goes to 0 at (0,0) to find a solution. or if the solution doesn't exist, try to find two paths with two different results in 0,0. whichever path i chose, this goes to +infinite. in class i have never seen a limit in two variables that exists and goes to +inf, having an indetermination form, so i don't have a clue on how to solve it. any ideas?

Please stop using bold font; it is distracting and it looks like you are yelling at us.

Anyway, your thoughts as given in your first two sentences are not correct. There are lots of other ways that may be used besides the two you mention. Sometimes a method that works in one problem fails in another problem, so you need to try something else.

If I were trying this problem I would switch to polar coordinates ##x = r \cos \theta,\: y = r \sin \theta,## with ## r > 0## and ##0 \leq \theta < 2 \pi## (in radians). Then, you want the limit as ##r \to 0+##. If the limit exists for any ##\theta,## and does not depend on the value of ##\theta,## then you have a genuine 2D limit. If the limit fails to exist for at least some values of ##\theta,## then your 2D limit does not exist. If the limit exists for all ##\theta## but is different for different values of ##\theta,## then again your 2D limit does not exist. This problem calls out for the use of l'Hospital's rule.

Finally: this message belongs in "Calculus and Beyond" rather than "Introductory Physics".

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i think what you are saying is incorrect, the polar form doesn't include non linear paths, so you can't be sure if the limits exist.

If you choose a path and it gives + infinity, then the limit can't exist (if it would exist, it would give the limit value along every path).
yea but, what if every path you find gives you +inf?

yea but, what if every path you find gives you +inf?

Does it matter? infinity is not a real number thus cannot be a limit by definition.

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i think what you are saying is incorrect, the polar form doesn't include non linear paths, so you can't be sure if the limits exist.

He never claims that this method proves that the limit exists. He rather claims that this method can be used to show that the limit does NOT exist, and he is completely right when saying this.

in R* the limit exist and I am considering that domain.
He never claims that this method proves that the limit exists. He rather claims that this method can be used to show that the limit does NOT exist, and he is completely right when saying this.
whatever, it doesn't work anyways, it goes to infinite for any angle

i think what you are saying is incorrect, the polar form doesn't include non linear paths, so you can't be sure if the limits exist.
The polar form does include nonlinear paths -- what's happening is that r is approaching zero, but ##\theta## isn't specified. A point (x, y) could be approaching the pole along a spiral or other path along some curve.

• Delta2
in R* the limit exist and I am considering that domain.

whatever, it doesn't work anyways

Remember to be polite to people who are trying to help you... Or even people in general...

So, if you claim that the limit is ##+\infty##, there is work to do. You have to prove:

##\forall M \in \mathbb{R} : \exists \delta > 0: \forall (x,y) \in \mathbb{R}^2: 0 < \Vert (x,y) \Vert < \delta \implies \frac{\ln(1+2x^2 + y^2)}{(x^2 + 3y^2)^2} > M##

Or you can settle with the easier answer that the limit doesn't exist in ##\mathbb{R}## (probably the answer your instructor intended).

in R* the limit exist and I am considering that domain.
What do you mean by R*? I'm not familiar with that notation. And how do you know the limit exists?

whatever, it doesn't work anyways, it goes to infinite for any angle
Which means the limit doesn't exist. So again, why do you think the limit exists?

What do you mean by R*? I'm not familiar with that notation. And how do you know the limit exists?

Which means the limit doesn't exist. So again, why do you think the limit exists?

I think he means the limit may be infinite.

The polar form does include nonlinear paths -- what's happening is that r is approaching zero, but θ\theta isn't specified. A point (x, y) could be approaching the pole along a spiral or other path along some curve.
it doesnt, that's why ((x^2)*y)/(x^4+y^2)) result in 0 in (0,0) with polar form for every angle but actually doesn't exist, i think its because of the way the distance is arranged.
I think he means the limit may be infinite.
yep.
Remember to be polite to people who are trying to help you... Or even people in general...
didnt mean to appear rude, sorry for that.
∀M∈R:∃δ>0:∀(x,y)∈R2:0<∥(x,y)∥<δ⟹ln(1+2x2+y2)(x2+3y2)2>M
is it possible to find a function that is always smaller that i know goes to infinite in 0,0?

i think what you are saying is incorrect, the polar form doesn't include non linear paths, so you can't be sure if the limits exist.
The polar representation sometimes allows you to say for sure that either that a limit exists, or that it does not.

Certainly, the polar form suffices in this case to draw a conclusion about the existence-or-not of a limit, and if you had actually tried it for yourself you would have seen this.

it doesnt, that's why ((x^2)*y)/(x^4+y^2)) result in 0 in (0,0) with polar form for every angle but actually doesn't exist, i think its because of the way the distance is arranged.

yep.

didnt mean to appear rude, sorry for that.

is it possible to find a function that is always smaller that i know goes to infinite in 0,0?
If you look at the graph of the ##\ln## function you will see that ##0 < \ln (1 + 2x^2 + y^2) < 2x^2+y^2 < 2(x^2+y^2) ##. Also: ##(x^2 + 3 y^2)^2 > (x^2+y^2)^2,## so the rest is pretty easy.

Certainly, the polar form suffices in this case to draw a conclusion about the existence-or-not of a limit, and if you had actually tried it for yourself you would have seen this.
you know what i do? do you have a camera in my room?i've spent hours on this limit. what you are saying is just incorrect. if id write in my exam "sometimes the polar form sufficies to draw conclusions" i would get a 0. it can suggest you what is the result, it can make you certain, but you would not prove it mathematically.

• Delta2
If you look at the graph of the ln\ln function you will see that 0<ln(1+2x2+y2)<2x2+y2<2(x2+y2)0 < \ln (1 + 2x^2 + y^2) < 2x^2+y^2 < 2(x^2+y^2) . Also: (x2+3y2)2>(x2+y2)2,(x^2 + 3 y^2)^2 > (x^2+y^2)^2, so the rest is pretty easy.
you are basically saying that the limit is smaller then +infinite, that doesn't give you information about the exact value.

you are basically saying that the limit is smaller then +infinite, that doesn't give you information about the exact value.

Right: sorry about that. Just forget post #15.

you know what i do? do you have a camera in my room?i've spent hours on this limit. what you are saying is just incorrect. if id write in my exam "sometimes the polar form sufficies to draw conclusions" i would get a 0. it can suggest you what is the result, it can make you certain, but you would not prove it mathematically.

Well, when I do it I get a definite conclusion, with no waffling about it working sometimes or not. In this case it actually does work, but I cannot write here the required steps to show that (not allowed under PF rules).

Well, when I do it I get a definite conclusion, with no waffling about it working sometimes or not. In this case it actually does work, but I cannot write here the required steps to show that (not allowed under PF rules).
cant i just say the function is > then f(x,y)=(ln(1+x)/(x^4)) that goes to +inf and so my function goes to +inf aswell?

The polar form does include nonlinear paths -- what's happening is that r is approaching zero, but ##\theta## isn't specified. A point (x, y) could be approaching the pole along a spiral or other path along some curve.

it doesnt, that's why ((x^2)*y)/(x^4+y^2)) result in 0 in (0,0) with polar form for every angle but actually doesn't exist, i think its because of the way the distance is arranged.
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.

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.

The polar form is very useful in this example, because the function becomes
$$f = \frac{\ln(1+r^2+r^2 \cos^2 \theta)}{r^4 (9 - 12 \cos^2 \theta + 4 \cos^4 \theta)},$$
which is of the form
$$\frac{\ln(1 + a r^2)}{b r^4}$$ for some "constants" ##a,b## that depend on ##\theta,## but are positive for all ##\theta \in [0,2 \pi].## So, the question is whether or not something of the form ##\ln(1+a r^2)/r^4## has a limit as ##r \to 0.## One way to answer that is to bound the logarithm in some reasonable way; another way is to use l'Hospital's rule.

The polar form is very useful in this example
The one I was referring to was ##f(x, y) = \frac{x^2y}{x^4 + y^2}##

The one I was referring to was ##f(x, y) = \frac{x^2y}{x^4 + y^2}##
Yes, I agree. That is a good example of how the polar method sometimes fails, but works often enough that it is still worth trying on an unexplored example.

it doesnt, that's why ((x^2)*y)/(x^4+y^2)) result in 0 in (0,0) with polar form for every angle but actually doesn't exist, i think its because of the way the distance is arranged.

yep.

didnt mean to appear rude, sorry for that.

is it possible to find a function that is always smaller that i know goes to infinite in 0,0?

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
$$f = \frac{\ln(1+2x^2 + y^2)}{(x^2 + 3 y^2)^2} > \frac{r^2 - \frac 1 2 r^4}{9 r^4}= \frac 1 9 \left(\frac{1}{r^2} - \frac{1}{2} \right)$$

• Kenneth1997 and member 587159
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 doesn't include linear paths, or the theta would be a variable not a constant, and in that case you couldn't 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'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 doesn't 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 don't have an idea on the best way to understand it. any thoughts?
hope you didnt register to this forum just for this

still think it doesn't 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.

thanks man, this should do the trick

still think it doesn't include linear paths, or the theta would be a variable not a constant, and in that case you couldn't 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.

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.

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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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)?

it wouldn't even be possible for the function i was trying to solve, and certainly it wouldn't help for finding a limit

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 doesn't include non linear paths, so you can't 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.

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.