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Showing that a limit of two variables doesn't exist

skatenerd

Active member
Oct 3, 2012
114
I had two of these problems assigned. I have to show that the limit doesn't exist for two separate functions as (x,y) approaches (0,0).
The first function was
$$\frac{x^4-y^2}{x^4+y^2}$$
and I went about showing the limit didn't exist by approaching along the x-axis to (0,0) and along the y-axis to (0,0) and showed that these two limits were different and therefore the limit as (x,y) approaches (0,0) doesn't exist. However I am kind of stumped with the next function.
The function is
$$\frac{x^2(y)}{x^4+y^2}$$
For this I tried the same thing as the first function, but seeing as the x and y on top multiply by each other you end up with 0 for both limits, proving inconclusive. I also tried subbing in \(y=kx^2\) where k is a constant and also \(y=kx\) and neither seemed to work out either. I think I'm missing something here...
 

chisigma

Well-known member
Feb 13, 2012
1,704
Re: showing that a limit of two variables doesn't exist

I had two of these problems assigned. I have to show that the limit doesn't exist for two separate functions as (x,y) approaches (0,0).
The first function was
$$\frac{x^4-y^2}{x^4+y^2}$$
and I went about showing the limit didn't exist by approaching along the x-axis to (0,0) and along the y-axis to (0,0) and showed that these two limits were different and therefore the limit as (x,y) approaches (0,0) doesn't exist. However I am kind of stumped with the next function.
The function is
$$\frac{x^2(y)}{x^4+y^2}$$
For this I tried the same thing as the first function, but seeing as the x and y on top multiply by each other you end up with 0 for both limits, proving inconclusive. I also tried subbing in \(y=kx^2\) where k is a constant and also \(y=kx\) and neither seemed to work out either. I think I'm missing something here...
In cases like these the best way is to put $\displaystyle x= \rho\ \cos \theta$, $\displaystyle y= \rho\ \sin \theta$ and to find the $\displaystyle \lim_{\rho \rightarrow 0} f(\rho, \theta)$ which is independent from $\theta$. For the first function is...

$\displaystyle f(\rho, \theta) = \frac{\rho^{4}\ \cos^{4} \theta - \rho^{4} \sin^{4} \theta}{\rho^{4}\ \cos^{4} \theta + \rho^{4} \sin^{4} \theta} = \frac{\cos^{4} \theta - \sin^{4} \theta}{\cos^{4} \theta + \sin^{4} \theta}$ (1)

... and the limit of course depends from $\theta$ so that the limit doesn't exist. For the second function is...


$\displaystyle f(\rho, \theta) = \frac{\rho^{3}\ \cos^{4} \theta \sin \theta}{\rho^{4}\ \cos^{4} \theta + \rho^{2} \sin^{2} \theta} = \frac{\rho\ \cos^{2} \theta \sin \theta}{\rho^{2}\ \cos^{4} \theta + \sin^{2} \theta}$ (2)

... and the limit is 0 independently from $\theta$...

Kind regards

$\chi$ $\sigma$
 
Last edited:

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
Re: showing that a limit of two variables doesn't exist

In other words, what you were "missing" is that the limit does exist and you were trying to show that it didn't!
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,703
Re: showing that a limit of two variables doesn't exist

I had two of these problems assigned. I have to show that the limit doesn't exist for two separate functions as (x,y) approaches (0,0).
The first function was
$$\frac{x^4-y^2}{x^4+y^2}$$
and I went about showing the limit didn't exist by approaching along the x-axis to (0,0) and along the y-axis to (0,0) and showed that these two limits were different and therefore the limit as (x,y) approaches (0,0) doesn't exist. However I am kind of stumped with the next function.
The function is
$$\frac{x^2(y)}{x^4+y^2}$$
For this I tried the same thing as the first function, but seeing as the x and y on top multiply by each other you end up with 0 for both limits, proving inconclusive. I also tried subbing in \(\color{red}{y=kx^2}\) where k is a constant and also \(y=kx\) and neither seemed to work out either. I think I'm missing something here...
Try again! I think you'll find that the answer depends on $k$.
 

skatenerd

Active member
Oct 3, 2012
114
Actually, yes I did try again and came out with what seems like a solid answer,
$$\frac{k}{1+k^2}$$ which seems to work right in showing the limit doesn't exist since it is different for different values of \(k\) . However, I put this original limit problem in wolframalpha to check and it says the limit exists and that it is zero, assuming that all the variables are real valued. So does that mean that this technique I used applies for more than just real valued numbers? Or is my answer actually just wrong and it doesn't prove the nonexistence of the limit?
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,703
Actually, yes I did try again and came out with what seems like a solid answer,
$$\frac{k}{1+k^2}$$ which seems to work right in showing the limit doesn't exist since it is different for different values of \(k\) . However, I put this original limit problem in wolframalpha to check and it says the limit exists and that it is zero, assuming that all the variables are real valued. So does that mean that this technique I used applies for more than just real valued numbers? Or is my answer actually just wrong and it doesn't prove the nonexistence of the limit?
No, it means that your answer is right and on this occasion it is wolframalpha that is actually just wrong. The function $\dfrac{x^2y}{x^4+y^2}$ has the strange property that if you approach the origin along any straight line then you get the limit $0$. But if you approach the origin along a parabolic path of the form $y=kx^2$ then the limit is $\dfrac k{1+k^2}$. This shows that the function does not have a limiting value at $(0,0).$
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Actually, yes I did try again and came out with what seems like a solid answer,
$$\frac{k}{1+k^2}$$ which seems to work right in showing the limit doesn't exist since it is different for different values of \(k\) . However, I put this original limit problem in wolframalpha to check and it says the limit exists and that it is zero, assuming that all the variables are real valued. So does that mean that this technique I used applies for more than just real valued numbers? Or is my answer actually just wrong and it doesn't prove the nonexistence of the limit?
I wonder what strategy will wolframalpha use to calculate the limit!
 

chisigma

Well-known member
Feb 13, 2012
1,704
Re: showing that a limit of two variables doesn't exist

... for the second function is...


$\displaystyle f(\rho, \theta) = \frac{\rho^{3}\ \cos^{4} \theta \sin \theta}{\rho^{4}\ \cos^{4} \theta + \rho^{2} \sin^{2} \theta} = \frac{\rho\ \cos^{2} \theta \sin \theta}{\rho^{2}\ \cos^{4} \theta + \sin^{2} \theta}$ (2)

... and the limit is 0 independently from $\theta$...
The conclusion is wrong because if $\rho$ tends to 0 along the trajectory...

$\displaystyle k\ \rho= \frac{\sin \theta}{\cos ^{2} \theta}$ (1)

... is...

$\displaystyle f(\rho, \theta) = \frac{1}{k}\ \frac{\sin^{2} \theta}{\rho^{2}\ \cos^{4} \theta + \sin^{2} \theta}$ (2)

... and the limit is $\displaystyle \frac{1}{k}$ so that the the function doesn't have limit in [0,0]. This example shows what types of 'traps' sometime there are in problems like this...

Kind regards

$\chi$ $\sigma$