Showing that a limit of two variables doesn't exist

In summary, the conversation discusses the process of showing that the limit does not exist for two separate functions as (x,y) approaches (0,0). The first function is given as $\frac{x^4-y^2}{x^4+y^2}$ and the limit is shown to not exist by approaching along the x-axis and y-axis and getting different limits. The conversation then moves on to discussing the second function, $\frac{x^2(y)}{x^4+y^2}$, and the difficulties in finding the limit using the same method. A new approach is suggested, using polar coordinates, and the resulting limit is found to be dependent on the constant k. The speaker then questions the validity of their answer, as
  • #1
skate_nerd
176
0
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...
 
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  • #2
Re: showing that a limit of two variables doesn't exist

skatenerd said:
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:
  • #3
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!
 
  • #4
Re: showing that a limit of two variables doesn't exist

skatenerd said:
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 $k$.
 
  • #5
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?
 
  • #6
skatenerd said:
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 $(0,0).$
 
  • #7
skatenerd said:
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!
 
  • #8
Re: showing that a limit of two variables doesn't exist

chisigma said:
... 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

$\chi$ $\sigma$
 

FAQ: Showing that a limit of two variables doesn't exist

How do you determine if a limit of two variables doesn't exist?

To determine if a limit of two variables doesn't exist, you need to check if the limit approaches different values when approaching the point from different directions.

What is the difference between a one-sided limit and a two-sided limit?

A one-sided limit only considers the values approaching a point from one direction, while a two-sided limit considers values approaching from both directions.

Can a limit of two variables not exist if both one-sided limits exist?

Yes, a limit of two variables can still not exist even if both one-sided limits exist. This can happen if the one-sided limits approach different values.

How can you graphically represent a limit of two variables?

A limit of two variables can be represented by a point on a graph where the two sides of the graph approach different values, indicating that the limit does not exist.

Are there any algebraic methods to prove that a limit of two variables doesn't exist?

Yes, there are algebraic methods such as using the squeeze theorem, finding a contradiction, or showing that the function approaches different values along different paths.

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