# Multivariable Limit Problem: Find Values of k That Make Limit Exist

• Rippling Hysteresis
In summary, In part (a), the author suggests looking into a parameterization because of all the squared terms. They consider 1-D calc and intuition from 1-D calc before settling on L'Hopital. In part (b), the author tries to simplify the expression by thinking about the two cases where z=0 and r goes to zero, vs r=0 and z goes to zero. For the former, they get z^k(0)/z^(2k)= 0/z^k. However, when z=r, the numerator becomes (1+r^2)-1= r^2, which as long as 2-k>0, the limit is r^(2-k
Rippling Hysteresis
Homework Statement
f(x,y,z) = z^k(e^(x^2+y^2) -1))/(x^2+y^2+z^2)^k for when (x,y,z) does not equal (0,0,0). f(x,y,z)=0 otherwise.
a) Find all the real, positive values of k where the function is continuous at the origin. Thus, find the values of k where the limit at (x,y,z) -> (0,0,0) equals zero.
b) Find the values of k where each of the partials, fx, fy, fz evaluated at (0,0,0) exist. Find the values of these partials
Relevant Equations
L'Hopital?
Parameterization?
fx(0,0) = lim t->0 = f(t,0) -f(0,0)/t
(a) I thought perhaps a parameterization would be the place to begin given all the squared terms.
x=rcos(u)sin(v)
y=rsin(u)sin(v)
z=rcos(v)

That would yield: r^k(cos(v))^k*(e^(r^2*(sin(v))^2))/(r^(2k))
Canceling a r^k at each level: (cos(v))^k*(e^(r^2*(sin(v))^2))/(r^(k))

I'm not sure how important the trig terms are, since they are bounded between (-1,1), but they may be necessary to keep. My focus is drawn to the e^(r^2)/r^k term. When I graph that, regardless of exponent, that terms since to tend toward infinity as r->0, which makes sense. So I'm not sure what form of analysis or technique I should use to further simplify this.

Intuition from 1-D Calc says maybe L'Hopital? But a little unsure how to apply with MV, or if it's even relevant. w.r.t. 'r' the top would become (cos(v))^k*(e^(r^2*(sin(v))^2))*(2r*(sin(v))^2) and the bottom would become (k-1)*r^(k-1). I'm not sure how that helps, because then it's still effectively (0)(infinity)/(infinity).

(b) I have a feeling this part would depend on part (a), and I should use the limit definition formula.

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For part (I) at least, it seems like it would be a simpler first step to use cylindrical coordinates instead. Then it might help to think about the two cases when z=0 and r goes to zero, vs r=0 and z goes to zero.

Ooh, yeah, thanks. Substituting those parameters gives me:

z^k(e^r^2 -1)/(r^2+z^2)^k

So for the limit (r=0, z->0) would be z^k(0)/z^(2k)= 0/z^k, since (e^(r^2)-1)=0 at r=0. As z->0, isn't this of the form 0/0? I would think L'Hopital, but am a little unsure because there isn't really a function left on top.

On the other hand, for the limit(r->0, z=0), we'd get (e^(r^2)-1)/r^(2k). I would imagine the exponential term minus 1 would more strongly go to zero, independent of k though. I graphed this function for some pretty large k values and it still seems to tend to infinity.

But my feeling is that isn't quite right, because I'm not finding k dependence.As z->0, we have .

For ##\frac{e^{r^2}-1}{r^{2k}}## you should try writing out the Taylor polynomial of the exponential. Remember, in the limit at r goes to zero only the smallest degree term matters. Or you can use L'hopital's rule, it's the same thing as using Taylor polynomial expansions.

For the case where r=0, you are correct. The numerator is just 0, so the limit is zero. This is actually pretty helpful, as if you want the limit to exist you need the limit to be zero when ##r\neq0## as well.

Thanks once again! The numerator then becomes (1+r^2)-1= r^2. So the expression reduces to r^2/r^(2k). In order for the limit to go to zero, then then numerator should have the higher degree. So, the expression equals r^(2-2k), and when k<1, the limit would equal zero (very small number to positive exponent. If that makes sense, I feel good about part (a) and might try exploring the limit definition of the derivative for part (b), trying similar techniques.

Office_Shredder said:
For part (I) at least, it seems like it would be a simpler first step to use cylindrical coordinates instead. Then it might help to think about the two cases when z=0 and r goes to zero, vs r=0 and z goes to zero.
Office_Shredder said:
For ##\frac{e^{r^2}-1}{r^{2k}}## you should try writing out the Taylor polynomial of the exponential. Remember, in the limit at r goes to zero only the smallest degree term matters. Or you can use L'hopital's rule, it's the same thing as using Taylor polynomial expansions.

For the case where r=0, you are correct. The numerator is just 0, so the limit is zero. This is actually pretty helpful, as if you want the limit to exist you need the limit to be zero when ##r\neq0## as well.
Wait, now I'm doubting myself and thinking I made a mistake with the z=0, r -> 0 part. Won't the numerator actually become 0(e^(r^2)-1)/r^(2k), which is again 0 and independent of k?

Whoops good point. I should have suggested thinking about the case where z=r! (Then you're going to want to use the Taylor polynomial expansion)

Office_Shredder said:
Whoops good point. I should have suggested thinking about the case where z=r! (Then you're going to want to use the Taylor polynomial expansion)
OK cool! So I see it this way then: If z=r, it simplifies to r^k(e^(r^2)-1)/(4r^(2k). Using the expansion e^(r^2) ≈ 1 +r^2, we get the expression to become r^2/(4r^(2k)= 1/4(r^(2-k), where as long as 2-k>0, the limit is zero.

Nice problem! I wish it would have been in my book when I took this class 20 yrs ago lol. I'll be around if u need any help with the next part.

I think you made a couple of errors. The denominator should only have a factor of 2, not 4 I think, and it should be raised to the power of k. I don't think this affects the final answer though. I also think r^2/(4r^(2k)= 1/4(r^(2-k) is also forgetting the r^k I'm the numerator on the left, but you canceled it out on the right.

So we have as long as k>2, the limit converges to 0 when r=0, z=0, or z=r. What about other choices of z and r? Any other choice involves both z and r being nonzero, so can be described as ##z=\alpha r## for some ##alpha##. If you can prove the limit goes to zero for any choice of ##\alpha## in a uniform way (in particular, something like if z and r are picked to be small, then you can't pick ##\alpha## adversely to make f big again), then you have finished proving the limit exists. This calculation should look pretty similar to the one you just did.

## 1. What is a multivariable limit problem?

A multivariable limit problem involves finding the value of a function at a specific point in a multi-dimensional space. In other words, it is the limit of a function as multiple variables approach a certain point.

## 2. How is a multivariable limit problem different from a single variable limit problem?

In a single variable limit problem, there is only one variable approaching a specific value. In a multivariable limit problem, there are multiple variables approaching a specific point, making it more complex.

## 3. What is the importance of finding values of k that make the limit exist?

When the limit of a multivariable function exists, it means that the function is continuous at that specific point. This is important in many areas of science, such as physics and engineering, where continuous functions are essential for modeling and predicting real-world phenomena.

## 4. What are the steps to solving a multivariable limit problem?

The first step is to plug in the given values for the variables and see if the limit exists. If it does not, then further analysis is needed. Next, try approaching the point from different directions to see if the limit changes. If it does, then the limit does not exist. If it does not change, then the limit may exist, but further analysis is still needed. Finally, algebraic manipulation and substitution may be used to find values of k that make the limit exist.

## 5. Are there any common mistakes to avoid when solving a multivariable limit problem?

One common mistake is assuming that the limit exists just because the function is continuous at the given point. Another mistake is not considering the behavior of the function as the variables approach the point from different directions. It is also important to check for any discontinuities or undefined points in the function. Lastly, it is important to carefully manipulate and substitute variables to find values of k that make the limit exist, as simple algebraic errors can lead to incorrect solutions.

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