A limit in two variables

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 doesnt 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 dont have a clue on how to solve it. any ideas?

member 587159

Homework Statement

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

The Attempt at a Solution

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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 doesnt 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 dont 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).

Ray Vickson
Homework Helper
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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 doesnt 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 dont 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 doesnt include non linear paths, so you cant 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?

member 587159
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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member 587159
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.

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 im 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 doesnt work anyways, it goes to infinite for any angle

Mark44
Mentor
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.
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
member 587159
in R* the limit exist and im considering that domain.

whatever, it doesnt 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).

vela
Staff Emeritus
Homework Helper
in R* the limit exist and im 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 doesnt 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?

member 587159
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, thats why ((x^2)*y)/(x^4+y^2)) result in 0 in (0,0) with polar form for every angle but actually doesnt 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?

Ray Vickson
Homework Helper
Dearly Missed
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.
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.

Ray Vickson
Homework Helper
Dearly Missed
it doesnt, thats why ((x^2)*y)/(x^4+y^2)) result in 0 in (0,0) with polar form for every angle but actually doesnt 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 doesnt give you information about the exact value.

Ray Vickson
Homework Helper
Dearly Missed
you are basically saying that the limit is smaller then +infinite, that doesnt give you information about the exact value.

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

Ray Vickson
Homework Helper
Dearly Missed
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?

Mark44
Mentor
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, thats why ((x^2)*y)/(x^4+y^2)) result in 0 in (0,0) with polar form for every angle but actually doesnt 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.

Ray Vickson
Homework Helper
Dearly Missed
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.

Mark44
Mentor
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}##

Ray Vickson
Homework Helper
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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.

Ray Vickson
Homework Helper
Dearly Missed
it doesnt, thats why ((x^2)*y)/(x^4+y^2)) result in 0 in (0,0) with polar form for every angle but actually doesnt 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