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

[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.

 

[Office_Shredder]

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.

 

[Rippling Hysteresis]

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 .

 

[Office_Shredder]

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.

 

[Rippling Hysteresis]

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.

 

[Rippling Hysteresis]

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.

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?

 

[Office_Shredder]

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)

 

[Rippling Hysteresis]

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.

 

[benorin]

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.

 

[Office_Shredder]

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.