# Is this a valid limit?

1. Sep 9, 2004

### StatusX

Sorry, I don't know how to write this besides just plain text, but it shouldn't be too hard to read:

lim k->0 (e^kt-a)/k = at

It seems strange to me, and I have no idea how to derive it. Here's how I got it. Starting with:

dv/dt = a+kv

Seperate variables and integrate to get:

dv/(a+kv) = dt

(1/k)*ln(a+kv) = t+c (I'll set c=0 from here on)

a+kv = e^(kt)

v = (e^kt-a)/k

which is the left side of the limit. But as k goes to 0 in the original equation:

dv/dt = a

v = at

Is this right? If so, is there a better way to derive it? If not, where did I go wrong? (I know there are multiple divide by zeros, but I don't think that's the problem because k varies continuously, and the original curve could get closer and closer to a straight line as k is smaller and smaller but still greater than 0.)

2. Sep 9, 2004

### Tide

I think you mistyped your original equation but I think you're looking for l'Hopital's Rule, i.e. the limit of the ratio is the same as the limit of the ratio of the derivatives when both numerator and denominator of the original ratio tend to zero (simplified version!)

3. Sep 9, 2004

### StatusX

I don't think I mistyped. I derive it below, is there a mistake there?

4. Sep 9, 2004

### Tide

I'm having trouble understanding your very first equation and wonder what v is.

5. Sep 10, 2004

### HallsofIvy

Staff Emeritus
Do you mean e^(kt-a)/k or (e^(kt)- a)/k??

In any case, I don't see that this has a limit. In the first case, the numerator approaches e^(-a) while the denominator goes to 0. In the second case, the numerator goes to 1-a while the denominator goes to 0.

Whatever the problem is, whether you are trying L'Hopital's rule or not, there is no point in differentiating with respect to t. The only parameter being varied is k.

You CAN show that lim(k->0)(e^(kt)- 1)/k= t by using L'Hopital's rule (differentiating with respect to k) and so you might intend showing that
lim(k->0)(e^(kat)-1)/k= at.

6. Sep 10, 2004

### uart

Either that or
lim k->0 (ae^kt-a)/k = at,
but certainly the limit as originally posted is not valid.

7. Sep 10, 2004

### StatusX

You're right, there should be an a in front of e^kt. I set c=0 thinking it wouldn't matter, but that's the difference. Now it works with l'hopitals rule. Still, if you go through my derivation, it seems like the limit should still equal at+c for some c. Is that true?

By the way, I'm sorry if this was confusing. a and t are just parameters, and v is only used in the derivation. I adapted this from a physics problem (a was acceleration(gravity), k was the drag coefficient, etc.) where I was looking at how ideal motion is acheived as k->0. Maybe in that context it makes more sense, but I did change a few things so don't thake that too seriously.

8. Sep 11, 2004

### matt grime

Erm, well, if the limit exists, then yes, trivially there is some c such that the limit l, equals at+c, c=l-at for instance.

9. Sep 11, 2004

### StatusX

Here's what I was trying to say. I learned a little tex code to make it easier to read. This was the original limit:

$$\lim_{k \rightarrow 0} \\ \frac{e^{kt}-a}{k} = at$$

However, I made a mistake in the derivation, not treating the constant of integration correctly, and thought it should be modified to this:

$$\lim_{k \rightarrow 0} \\ \frac{C_{1}e^{kt}-a}{k} = at+C_{2}$$

But I didn't really think this through either, assuming the constants were completely arbitrary and that whatever I pick for $$C_{1}$$ (eg., 1, to get back my original limit), I could find another number for $$C_{2}$$ that would make the limit valid. Now I realize what the problem is. Here is my work:

$$\frac{dv}{dt} = k v+a$$

$$\frac{dv}{k v+a} = dt$$

Last time, I did two indefinite integrals with a constant, but since that didn't seem to work right, I tried it like this:

$$\int_{v_{0}}^{v} \frac{dv'}{k v'+a} = \int_{0}^{t} dt'$$

where $$v=v(t)$$ and $$v_{0}=v(0)$$

$$\frac{1}{k} (\ln(k v'+a)\mid_{v_{0}}^{v})= t$$

$$\ln(k v+a)-\ln(k v_{0}+a)= k t$$

$$\ln(\frac{k v+a}{k v_{0}+a})= k t$$

$$\frac{k v+a}{k v_{0}+a}= e^{k t}$$

$$k v+a=(k v_{0}+a) e^{k t}$$

$$v(t)=\frac{(k v_{0}+a) e^{k t}-a}{k}$$

Now you can see why the constant is not completely arbitrary. It depends not only on the initial conditions, but also on k, so it cannot be treated as an independent parameter in the limit.

$$\lim_{k \rightarrow 0} \ \frac{(k v_{0}+a) e^{k t}-a}{k}$$

$$= \lim_{k \rightarrow 0} \ \frac{v_{0} (k e^{k t})+a (e^{k t}-1)}{k}$$

This is 0/0, so by l'hopitals rule, you can differentiate the top and bottom to get:

$$=\lim_{k \rightarrow 0} \ \frac{v_{0} (k t e^{k t}+e^{k t})+a(t e^{k t})}{1}$$

$$= v_{0} (0+1)+a(t (1))$$

$$= at+v_{0}$$

So replacing $$v_{0}$$ with $$C$$, the limit I was looking for is:

$$\lim_{k \rightarrow 0} \ \frac{(k C + a) e^{k t}-a}{k} = at+C$$

This is valid, right? Thanks for the help everyone.

Last edited: Sep 11, 2004
10. Sep 12, 2004

### matt grime

The whole premise is wrong. Asking what the orginal limit is is meaningless unless a=1, obviously the limit doesn't exist unless a=1. Forget about finding these constants which is dubious at best.