# Finding the limit

1. Aug 1, 2009

### azizz

1. The problem statement, all variables and given/known data

For the function:

$$f(i,z) := \frac{1}{z-z_0} \left( \sqrt{g} + \sqrt{\frac{k}{m}} \frac{i-i_0}{z-z_0} \right)$$

we have to find a solution for $$f(i_0,z_0)$$.

2. Relevant equations

$$z_0 = \sqrt{\frac{k}{mg}} i_0$$

$$i_0 = 1$$

3. The attempt at a solution

As you can see the system goes to $$\infty$$ if we let $$z$$ and $$i$$ go to $$z_0$$ and $$i_0$$ at the same time.

So my fist guess was to first let $$i$$ go to $$i_0$$, this gives

$$f(i_0,z) := \frac{1}{z-z_0} \sqrt{g}$$

But again for $$z \rightarrow z_0$$ the function $$f(i_0,z_0)$$ will go to $$\infty$$ (even if I fill in the equation for $$z_0$$ as given above; no cancellations take place).

4. Hints from the professor

I asked my professor for a couple of hints, but Im still unable to solve the problem with this information. Perhaps you can use it?

- it you let both go to its nominal value ($$z_0$$ or $$i_0$$) at the same time then the function will go to $$\infty$$. But there exists a certain trajectory (fixed $$i=i_0$$ and $$z(i) \rightarrow z_0$$ or vv) where the function has a solution.

- so you cannot let $$z$$ and $$i$$ go to $$z_0$$ and $$i_0$$ at the same time. Instead you have to let one of the two go to its nominal value ($$z_0$$ or $$i_0$$) and then let the other go to its nominal value as a function of the other.

Anyone has a suggestion how this problem can be tackled?

Last edited: Aug 1, 2009
2. Aug 1, 2009

### Dick

How about making sure that sqrt(k/m)*(i-i0)/(z-z0) is always equal to -sqrt(g)?

3. Aug 1, 2009

### azizz

Thanks for the fast reply! That might be a solution indeed, because then f(i0,z0)=0. I havent checked this yet, but in the remainder of the exercise I have to let f(i0,z0) vary 50% from its nominal value (that is f(i0,z0)). So I dont expect this to be the solutions, but I will take a look at it.

4. Aug 1, 2009

### Dick

I rather suspect if you are clever about the path, you can make it converge to anything you want. That's what makes the problem not very interesting.

5. Aug 2, 2009

### azizz

I can see f(i0,z0)=0 is a solution indeed. I cant see that other values are solutions as well, but I can imagine your right. However, for the remainder of the exercise I expect only one solution to be the desired one. Is it possible that there exist something like a "best" solution (I doubt this though).

6. Aug 2, 2009

### Dick

Think about solving sqrt(g)+sqrt(k/m)*(i-i0)/(z-z0)=(z-z0)/K for i as a function of z and let i->i0. Wouldn't that give you a limit of K?

7. Aug 2, 2009

### azizz

Im sorry, but I still dont see how this is to be done. First of all, why introduce the new variable K? Isnt that the same as 1/f(i,z)?

Then you say that the limit of K can be found by solving the given equation for i as function of z, that would be

$$i = \sqrt{\frac{m}{k}} (z-z_0) \left( \frac{z-z_0}{K} - \sqrt{g} \right) + i_0$$

So for $$z \rightarrow z_0$$ we get $$i \rightarrow i_0$$, but we knew that already...

Last edited: Aug 2, 2009
8. Aug 2, 2009

### Dick

The point is to make limit of f(i,z) as i->i0, z->z0 equal to 1/K. Where K is pretty much anything we want. Your function f just plain doesn't have a definite limit. I'm not sure I can think of any interesting questions to ask about the 'limit'.

9. Aug 3, 2009

### HallsofIvy

Staff Emeritus
You say "limit" in the title but then you say just "find f(i0,z0)" in the body of you post. That value is, of course, NOT defined. Do you mean to define f(i0,z0) as the limit so that the function is continuous? (And again, as you have been told, that limit does not exist so this is impossible.)

10. Aug 16, 2009

### azizz

yes that is right, the limit does not exist. I misused this term. Indeed Im looking for a value for f(i0,z0) such that the function is continuous at this point.